52 results found for "triangle-circles" in Class 10.
सेनेगाम्बिया के पत्थर वृत्त किस पुरातात्विक परंपरा का प्रमाण हैं?
Stone Circles of Senegambia are evidence of which archaeological tradition?
#senegambia
#stone-circles
#megalithic
A मुगल शाही समाधियां / Mughal royal tombs
B पश्चिमी अफ्रीकी मेगालिथिक समाधि परंपरा / West African megalithic funerary tradition
C रोमन स्नानागार / Roman baths
D माया लकड़ी महल / Maya wooden palaces
Explanation opens after your attempt
Correct Answer
B. पश्चिमी अफ्रीकी मेगालिथिक समाधि परंपरा / West African megalithic funerary tradition
Step 1
Concept
These stone circles show ancient funerary and monument tradition of West Africa. For exams remember Gambia and Senegal.
Step 2
Why this answer is correct
The correct answer is B. पश्चिमी अफ्रीकी मेगालिथिक समाधि परंपरा / West African megalithic funerary tradition. These stone circles show ancient funerary and monument tradition of West Africa. For exams remember Gambia and Senegal.
Step 3
Exam Tip
ये पत्थर वृत्त पश्चिम अफ्रीका की प्राचीन समाधि और स्मारक परंपरा दिखाते हैं। परीक्षा में गाम्बिया और सेनेगल याद रखें।
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किसी पोस्टर में त्रिभुज का उपयोग खतरे या चेतावनी जैसा प्रभाव क्यों दे सकता है?
Why can use of a triangle in a poster give a danger or warning-like effect?
#triangle
#sharp angles
#poster
A क्योंकि उसमें नुकीले कोण होते हैं / Because it has sharp angles
B क्योंकि वह हमेशा नीला होता है / Because it is always blue
C क्योंकि वह छूने से खुरदरा होता है / Because it is rough to touch
D क्योंकि वह केवल ऋणात्मक स्थान है / Because it is only negative space
Explanation opens after your attempt
Correct Answer
A. क्योंकि उसमें नुकीले कोण होते हैं / Because it has sharp angles
Step 1
Concept
Sharp angles can suggest sharpness and caution. Exam tip: understand shape mood through context.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि उसमें नुकीले कोण होते हैं / Because it has sharp angles. Sharp angles can suggest sharpness and caution. Exam tip: understand shape mood through context.
Step 3
Exam Tip
नुकीले कोण तीखापन और सावधानी का संकेत दे सकते हैं। परीक्षा में shape mood को संदर्भ से समझें।
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वृत्त और त्रिभुज में समानता किस आधार पर है?
On what basis are circle and triangle similar?
#shape
#circle
#triangle
A दोनों द्वि आयामी आकार हैं / Both are two-dimensional shapes
B दोनों त्रि आयामी रूप हैं / Both are three-dimensional forms
C दोनों वास्तविक बनावट हैं / Both are actual textures
D दोनों रंग ताप हैं / Both are colour temperatures
Explanation opens after your attempt
Correct Answer
A. दोनों द्वि आयामी आकार हैं / Both are two-dimensional shapes
Step 1
Concept
Circle and triangle are both flat shapes. Exam tip: write flat closed area as shape.
Step 2
Why this answer is correct
The correct answer is A. दोनों द्वि आयामी आकार हैं / Both are two-dimensional shapes. Circle and triangle are both flat shapes. Exam tip: write flat closed area as shape.
Step 3
Exam Tip
वृत्त और त्रिभुज दोनों समतल आकार हैं। परीक्षा में flat closed area को shape लिखें।
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एक समकोण त्रिभुज का क्षेत्रफल (84) वर्ग सेमी है और एक लम्ब पक्ष दूसरे से (5) सेमी अधिक है। छोटे लम्ब पक्ष की लंबाई क्या है?
The area of a right triangle is (84) square cm and one perpendicular side is (5) cm more than the other. What is the shorter perpendicular side?
#quadratic equations
#right triangle
#area
A (7) सेमी / (7) cm
B (8) सेमी / (8) cm
C (12) सेमी / (12) cm
D (14) सेमी / (14) cm
Explanation opens after your attempt
Correct Answer
C. (12) सेमी / (12) cm
Step 1
Concept
Let the shorter side be (x). Then (\frac{1}{2}x(x+5)=84). This gives \(x^2+5x-168=0\), so (x=12).
Step 2
Why this answer is correct
The correct answer is C. (12) सेमी / (12) cm. Let the shorter side be (x). Then (\frac{1}{2}x(x+5)=84). This gives \(x^2+5x-168=0\), so (x=12).
Step 3
Exam Tip
छोटा पक्ष (x) हो तो (\frac{1}{2}x(x+5)=84)। इससे \(x^2+5x-168=0\) और (x=12) है।
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एक समकोण त्रिभुज के दो छोटे पक्ष (x) सेमी और (x+2) सेमी हैं तथा कर्ण (10) सेमी है। बड़ा छोटा पक्ष क्या है?
The two shorter sides of a right triangle are (x) cm and (x+2) cm, and the hypotenuse is (10) cm. What is the larger shorter side?
#quadratic equations
#right triangle
#application
A (6) सेमी / (6) cm
B (7) सेमी / (7) cm
C (8) सेमी / (8) cm
D (9) सेमी / (9) cm
Explanation opens after your attempt
Correct Answer
C. (8) सेमी / (8) cm
Step 1
Concept
We get (x-2 +(x+2)2 =100). This gives (x=6), so the larger shorter side is (8) cm.
Step 2
Why this answer is correct
The correct answer is C. (8) सेमी / (8) cm. We get (x-2 +(x+2)2 =100). This gives (x=6), so the larger shorter side is (8) cm.
Step 3
Exam Tip
(x-2 +(x+2)2 =100) बनता है। इससे (x=6) मिलता है इसलिए बड़ा छोटा पक्ष (8) सेमी है।
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एक समकोण त्रिभुज में लम्ब आधार से (7) सेमी अधिक है और कर्ण (25) सेमी है। आधार कितना है?
In a right triangle, the perpendicular is (7) cm more than the base and the hypotenuse is (25) cm. What is the base?
#quadratic equations
#right triangle
#pythagoras
A (15) सेमी / (15) cm
B (16) सेमी / (16) cm
C (18) सेमी / (18) cm
D (20) सेमी / (20) cm
Explanation opens after your attempt
Correct Answer
A. (15) सेमी / (15) cm
Step 1
Concept
The base is (x) and perpendicular is (x+7). From (x-2 +(x+7)2 =252 ), (x=15).
Step 2
Why this answer is correct
The correct answer is A. (15) सेमी / (15) cm. The base is (x) and perpendicular is (x+7). From (x-2 +(x+7)2 =252 ), (x=15).
Step 3
Exam Tip
आधार (x) और लम्ब (x+7) है। (x-2 +(x+7)2 =252 ) से (x=15) मिलता है।
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एक समकोण त्रिभुज का कर्ण (17,सेमी) है और एक भुजा दूसरी से (7,सेमी) अधिक है। छोटी भुजा क्या है?
The hypotenuse of a right triangle is (17,cm) and one side is (7,cm) more than the other. What is the smaller side?
#quadratic equations
#right triangle
#pythagoras
A (6,सेमी) / (6,cm)
B (7,सेमी) / (7,cm)
C (8,सेमी) / (8,cm)
D (9,सेमी) / (9,cm)
Explanation opens after your attempt
Correct Answer
C. (8,सेमी) / (8,cm)
Step 1
Concept
If the smaller side is (x), then (x-2 +(x+7)2 =172 ). This gives (x=8).
Step 2
Why this answer is correct
The correct answer is C. (8,सेमी) / (8,cm\(). If the smaller side is (x), then (x^2+(x+7)^2=17^2). This gives (x=8).\)
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+7)2 =172 )। इससे (x=8) मिलता है।
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एक त्रिभुज का क्षेत्रफल \(60,सेमी^2\) है। उसका आधार ऊँचाई से (4,सेमी) अधिक है। ऊँचाई क्या है?
The area of a triangle is (60,cm\(^2). Its base is (4\),cm) more than its height. What is the height?
#quadratic equations
#triangle area
#application
A (8,सेमी) / (8,cm)
B (10,सेमी) / (10,cm)
C (12,सेमी) / (12,cm)
D (14,सेमी) / (14,cm)
Explanation opens after your attempt
Correct Answer
B. (10,सेमी) / (10,cm)
Step 1
Concept
If height is (x), then (\frac{1}{2}x(x+4)=60). Hence (x=10).
Step 2
Why this answer is correct
The correct answer is B. (10,सेमी) / (10,cm\(). If height is (x), then (\frac{1}{2}x(x+4)=60). Hence (x=10).\)
Step 3
Exam Tip
यदि ऊँचाई (x) है तो (\frac{1}{2}x(x+4)=60)। इसलिए (x=10) है।
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एक समकोण त्रिभुज में छोटी भुजा (x cm), दूसरी भुजा (x+7 cm) और कर्ण (x+8 cm) है। छोटी भुजा क्या है?
In a right triangle, the shorter side is (x cm), the other side is (x+7 cm), and the hypotenuse is (x+8 cm). What is the shorter side?
#quadratic equations
#right triangle
#pythagoras
A (5 cm)
B (6 cm)
C (7 cm)
D (8 cm)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+7)2 =(x+8)2 ). This gives \(x^2-2x-15=0\), so (x=5).
Step 2
Why this answer is correct
The correct answer is A. (5 cm\(). By Pythagoras, (x^2+(x+7)^2=(x+8)^2). This gives (x^2-2x-15=0), so (x=5).\)
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+7)2 =(x+8)2 )। इससे \(x^2-2x-15=0\), इसलिए (x=5)।
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एक त्रिभुज का आधार ऊँचाई से (6 cm) अधिक है और क्षेत्रफल \(140 cm^2\) है। ऊँचाई क्या है?
The base of a triangle is (6 cm) more than its height, and its area is (140 cm\(^2). What is the height\)?
#quadratic equations
#triangle area
#application
A (10 cm)
B (14 cm)
C (16 cm)
D (20 cm)
Explanation opens after your attempt
Correct Answer
B. (14 cm)
Step 1
Concept
Let height be (x), then (\frac{1}{2}x(x+6)=140). This gives \(x^2+6x-280=0\), so (x=14).
Step 2
Why this answer is correct
The correct answer is B. (14 cm\(). Let height be (x), then (\frac{1}{2}x(x+6)=140). This gives (x^2+6x-280=0), so (x=14).\)
Step 3
Exam Tip
ऊँचाई (x) हो, तो (\frac{1}{2}x(x+6)=140)। इससे \(x^2+6x-280=0\), इसलिए (x=14)।
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एक समकोण त्रिभुज में कर्ण (185) सेमी है और एक लम्ब भुजा दूसरी से (37) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (185) cm and one leg is (37) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (96) सेमी / (96) cm
B (111) सेमी / (111) cm
C (120) सेमी / (120) cm
D (148) सेमी / (148) cm
Explanation opens after your attempt
Correct Answer
B. (111) सेमी / (111) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+37)2 =1852 ). This gives (x=111).
Step 2
Why this answer is correct
The correct answer is B. (111) सेमी / (111) cm. If the smaller leg is (x), then (x-2 +(x+37)2 =1852 ). This gives (x=111).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+37)2 =1852 )। इससे (x=111) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+30) सेमी हैं तथा कर्ण (150) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+30) cm and the hypotenuse is (150) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (75) सेमी / (75) cm
B (80) सेमी / (80) cm
C (90) सेमी / (90) cm
D (120) सेमी / (120) cm
Explanation opens after your attempt
Correct Answer
C. (90) सेमी / (90) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+30)2 =1502 ), giving (x=90). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is C. (90) सेमी / (90) cm. By Pythagoras, (x-2 +(x+30)2 =1502 ), giving (x=90). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+30)2 =1502 ), जिससे (x=90) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (145) सेमी है और एक लम्ब भुजा दूसरी से (17) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (145) cm and one leg is (17) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (84) सेमी / (84) cm
B (88) सेमी / (88) cm
C (95) सेमी / (95) cm
D (112) सेमी / (112) cm
Explanation opens after your attempt
Correct Answer
A. (84) सेमी / (84) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+17)2 =1452 ). This gives (x=84).
Step 2
Why this answer is correct
The correct answer is A. (84) सेमी / (84) cm. If the smaller leg is (x), then (x-2 +(x+17)2 =1452 ). This gives (x=84).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+17)2 =1452 )। इससे (x=84) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+24) सेमी हैं तथा कर्ण (120) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+24) cm and the hypotenuse is (120) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (66) सेमी / (66) cm
B (72) सेमी / (72) cm
C (80) सेमी / (80) cm
D (96) सेमी / (96) cm
Explanation opens after your attempt
Correct Answer
B. (72) सेमी / (72) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+24)2 =1202 ), giving (x=72). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (72) सेमी / (72) cm. By Pythagoras, (x-2 +(x+24)2 =1202 ), giving (x=72). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+24)2 =1202 ), जिससे (x=72) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (109) सेमी है और एक लम्ब भुजा दूसरी से (19) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (109) cm and one leg is (19) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (60) सेमी / (60) cm
B (65) सेमी / (65) cm
C (72) सेमी / (72) cm
D (84) सेमी / (84) cm
Explanation opens after your attempt
Correct Answer
A. (60) सेमी / (60) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+19)2 =1092 ). This gives (x=60).
Step 2
Why this answer is correct
The correct answer is A. (60) सेमी / (60) cm. If the smaller leg is (x), then (x-2 +(x+19)2 =1092 ). This gives (x=60).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+19)2 =1092 )। इससे (x=60) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+18) सेमी हैं तथा कर्ण (90) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+18) cm and the hypotenuse is (90) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (48) सेमी / (48) cm
B (54) सेमी / (54) cm
C (60) सेमी / (60) cm
D (72) सेमी / (72) cm
Explanation opens after your attempt
Correct Answer
B. (54) सेमी / (54) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+18)2 =902 ), giving (x=54). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (54) सेमी / (54) cm. By Pythagoras, (x-2 +(x+18)2 =902 ), giving (x=54). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+18)2 =902 ), जिससे (x=54) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (41) है और एक भुजा दूसरी से (31) अधिक है। छोटी भुजा क्या है?
In a right triangle, the hypotenuse is (41) and one leg is (31) more than the other. What is the smaller leg?
#quadratic-equations
#word-problems
#right-triangle
A (9)
B (40)
C (31)
D (10)
Explanation opens after your attempt
Step 1
Concept
If the smaller leg is (x), the other is (x+31). From (x-2 +(x+31)2 =412 ), (x=9).
Step 2
Why this answer is correct
The correct answer is A. (9). If the smaller leg is (x), the other is (x+31). From (x-2 +(x+31)2 =412 ), (x=9).
Step 3
Exam Tip
यदि छोटी भुजा (x) है, तो दूसरी (x+31) होगी। (x-2 +(x+31)2 =412 ) से (x=9) मिलता है।
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एक समकोण त्रिभुज में कर्ण (85) सेमी है और एक लम्ब भुजा दूसरी से (13) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (85) cm and one leg is (13) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (48) सेमी / (48) cm
B (51) सेमी / (51) cm
C (60) सेमी / (60) cm
D (72) सेमी / (72) cm
Explanation opens after your attempt
Correct Answer
B. (51) सेमी / (51) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+13)2 =852 ). This gives (x=51).
Step 2
Why this answer is correct
The correct answer is B. (51) सेमी / (51) cm. If the smaller leg is (x), then (x-2 +(x+13)2 =852 ). This gives (x=51).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+13)2 =852 )। इससे (x=51) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+14) सेमी हैं तथा कर्ण (70) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+14) cm and the hypotenuse is (70) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (42) सेमी / (42) cm
B (44) सेमी / (44) cm
C (46) सेमी / (46) cm
D (48) सेमी / (48) cm
Explanation opens after your attempt
Correct Answer
A. (42) सेमी / (42) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+14)2 =702 ), giving (x=42). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is A. (42) सेमी / (42) cm. By Pythagoras, (x-2 +(x+14)2 =702 ), giving (x=42). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+14)2 =702 ), जिससे (x=42) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (34) है और एक भुजा दूसरी से (14) अधिक है। छोटी भुजा क्या है?
In a right triangle, the hypotenuse is (34) and one side is (14) more than the other. What is the smaller side?
#quadratic-equations
#word-problems
#right-triangle
A (16)
B (30)
C (18)
D (20)
Explanation opens after your attempt
Step 1
Concept
If the smaller side is (x), the other side is (x+14). From (x-2 +(x+14)2 =342 ), (x=16).
Step 2
Why this answer is correct
The correct answer is A. (16). If the smaller side is (x), the other side is (x+14). From (x-2 +(x+14)2 =342 ), (x=16).
Step 3
Exam Tip
यदि छोटी भुजा (x) है, तो दूसरी (x+14) होगी। (x-2 +(x+14)2 =342 ) से (x=16) मिलता है।
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एक समकोण त्रिभुज में कर्ण (17) है और एक भुजा दूसरी से (7) अधिक है। छोटी भुजा क्या है?
In a right triangle, the hypotenuse is (17) and one leg is (7) more than the other. What is the smaller leg?
#quadratic-equations
#word-problems
#right-triangle
A (8)
B (15)
C (10)
D (7)
Explanation opens after your attempt
Step 1
Concept
If the smaller leg is (x), the other is (x+7). From (x-2 +(x+7)2 =172 ), (x=8).
Step 2
Why this answer is correct
The correct answer is A. (8). If the smaller leg is (x), the other is (x+7). From (x-2 +(x+7)2 =172 ), (x=8).
Step 3
Exam Tip
यदि छोटी भुजा (x) है, तो दूसरी (x+7) होगी। (x-2 +(x+7)2 =172 ) से (x=8) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+12) सेमी हैं तथा कर्ण (60) सेमी है। (x) का मान क्या है?
The legs of a right triangle are (x) cm and (x+12) cm, and the hypotenuse is (60) cm. What is the value of (x)?
#quadratic equations
#pythagoras
#right triangle
A (30)
B (36)
C (42)
D (48)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+12)2 =602 ), giving (x=36). Always take the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (36). By Pythagoras, (x-2 +(x+12)2 =602 ), giving (x=36). Always take the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+12)2 =602 ), जिससे (x=36) है। कर्ण को हमेशा सबसे बड़ी भुजा मानें।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+8) सेमी हैं तथा कर्ण (40) सेमी है। (x) का मान क्या है?
The legs of a right triangle are (x) cm and (x+8) cm, and the hypotenuse is (40) cm. What is the value of (x)?
#quadratic equations
#pythagoras
#right triangle
A (20)
B (24)
C (28)
D (32)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+8)2 =402 ), giving (x=24). In a right triangle, forming the correct equation is most important.
Step 2
Why this answer is correct
The correct answer is B. (24). By Pythagoras, (x-2 +(x+8)2 =402 ), giving (x=24). In a right triangle, forming the correct equation is most important.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+8)2 =402 ), जिससे (x=24) है। समकोण त्रिभुज में सही समीकरण बनाना सबसे जरूरी है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+5) सेमी हैं तथा कर्ण (25) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+5) cm, and the hypotenuse is (25) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (12) सेमी / (12) cm
B (15) सेमी / (15) cm
C (18) सेमी / (18) cm
D (20) सेमी / (20) cm
Explanation opens after your attempt
Correct Answer
B. (15) सेमी / (15) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+5)2 =252 ), giving (x=15). Always take the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (15) सेमी / (15) cm. By Pythagoras, (x-2 +(x+5)2 =252 ), giving (x=15). Always take the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+5)2 =252 ), जिससे (x=15) है। कर्ण को हमेशा सबसे बड़ी भुजा मानें।
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एक समकोण त्रिभुज में लम्ब भुजाएँ (x) सेमी और (x+1) सेमी हैं तथा कर्ण (5) सेमी है। छोटी भुजा क्या है?
A right triangle has legs (x) cm and (x+1) cm and hypotenuse (5) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (2) सेमी / (2) cm
B (3) सेमी / (3) cm
C (4) सेमी / (4) cm
D (5) सेमी / (5) cm
Explanation opens after your attempt
Correct Answer
B. (3) सेमी / (3) cm
Step 1
Concept
(x-2 +(x+1)2 =52 ) gives (x=3). After solving, check the (3,4,5) triplet.
Step 2
Why this answer is correct
The correct answer is B. (3) सेमी / (3) cm. (x-2 +(x+1)2 =52 ) gives (x=3). After solving, check the (3,4,5) triplet.
Step 3
Exam Tip
(x-2 +(x+1)2 =52 ) से (x=3) मिलता है। हल के बाद (3,4,5) त्रिक की जाँच करें।
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एक समकोण त्रिभुज का आधार (x+5), ऊंचाई (x+9) और क्षेत्रफल (95) है। सही समीकरण कौन-सा है?
A right triangle has base (x+5), height (x+9), and area (95). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#expert
A \(x^2+14x-145=0\)
B \(x^2+14x-95=0\)
C \(x^2+14x+45=0\)
D \(x^2+14x-190=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+14x-145=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+5)(x+9)=95). Thus ((x+5)(x+9)=190) and \(x^2+14x-145=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+14x-145=0\). The area is (\frac{1}{2}(x+5)(x+9)=95). Thus ((x+5)(x+9)=190) and \(x^2+14x-145=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+5)(x+9)=95) होगा। इसलिए ((x+5)(x+9)=190) और \(x^2+14x-145=0\)।
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एक समकोण त्रिभुज का आधार (x+4), ऊंचाई (x+8) और क्षेत्रफल (72) है। सही समीकरण कौन-सा है?
A right triangle has base (x+4), height (x+8), and area (72). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#expert
A \(x^2+12x-112=0\)
B \(x^2+12x-72=0\)
C \(x^2+12x+32=0\)
D \(x^2+12x-144=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+12x-112=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+12x-112=0\). The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+4)(x+8)=72) होगा। इसलिए ((x+4)(x+8)=144) और \(x^2+12x-112=0\)।
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एक समकोण त्रिभुज का आधार (x+3), ऊंचाई (x+7) और क्षेत्रफल (55) है। सही समीकरण कौन-सा है?
A right triangle has base (x+3), height (x+7), and area (55). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#expert
A \(x^2+10x-89=0\)
B \(x^2+10x-55=0\)
C \(x^2+10x+21=0\)
D \(x^2+10x-110=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+10x-89=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+3)(x+7)=55). Thus ((x+3)(x+7)=110) and \(x^2+10x-89=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+10x-89=0\). The area is (\frac{1}{2}(x+3)(x+7)=55). Thus ((x+3)(x+7)=110) and \(x^2+10x-89=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+3)(x+7)=55) होगा। इसलिए ((x+3)(x+7)=110) और \(x^2+10x-89=0\)।
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एक समकोण त्रिभुज का आधार (x+2), ऊंचाई (x+6) और क्षेत्रफल (40) है। सही समीकरण कौन-सा है?
A right triangle has base (x+2), height (x+6), and area (40). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#hard
A \(x^2+8x-68=0\)
B \(x^2+8x-40=0\)
C \(x^2+8x+12=0\)
D \(x^2+8x-80=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+8x-68=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+2)(x+6)=40). Thus ((x+2)(x+6)=80) and \(x^2+8x-68=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+8x-68=0\). The area is (\frac{1}{2}(x+2)(x+6)=40). Thus ((x+2)(x+6)=80) and \(x^2+8x-68=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+2)(x+6)=40) होगा। इसलिए ((x+2)(x+6)=80) और \(x^2+8x-68=0\)।
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एक समकोण त्रिभुज का आधार (x+1), ऊंचाई (x+3) और क्षेत्रफल (30) है। सही समीकरण कौन-सा है?
A right triangle has base (x+1), height (x+3), and area (30). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#hard
A \(x^2+4x-57=0\)
B \(x^2+4x-30=0\)
C \(x^2+4x+3=0\)
D \(x^2+4x-60=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+4x-57=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+4x-57=0\). The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+1)(x+3)=30) होगा। इसलिए ((x+1)(x+3)=60) और \(x^2+4x-57=0\)।
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एक समकोण त्रिभुज का आधार (x), ऊँचाई (x+2) और क्षेत्रफल (24) है। सही द्विघात समीकरण कौन-सा है?
A right triangle has base (x), height (x+2), and area (24). Which quadratic equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#medium
A \(x^2+2x-48=0\)
B \(x^2+2x-24=0\)
C \(2x^2+2x-24=0\)
D \(x^2-2x-48=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x-48=0\)
Step 1
Concept
The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x-48=0\). The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}x(x+2)=24) होगा। इसलिए (x(x+2)=48) और \(x^2+2x-48=0\) मिलता है।
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यदि किसी चित्र में केवल वृत्त और वर्ग हैं तो कौन सा तत्व प्रमुख है?
If a picture has only circles and squares which element is prominent?
#shape
#circle
#square
A बनावट / Texture
B आकार / Shape
C स्थान / Space
D मान / Value
Explanation opens after your attempt
Correct Answer
B. आकार / Shape
Step 1
Concept
Circles and squares are examples of shapes. Exam tip: write flat closed figures as shapes.
Step 2
Why this answer is correct
The correct answer is B. आकार / Shape. Circles and squares are examples of shapes. Exam tip: write flat closed figures as shapes.
Step 3
Exam Tip
वृत्त और वर्ग आकार के उदाहरण हैं। परीक्षा में flat closed figures को shapes लिखें।
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वृत्त वर्ग और त्रिभुज किसके उदाहरण हैं?
Circle square and triangle are examples of what?
#shape
#geometric shapes
#circle square
A बनावट / Texture
B आकार / Shape
C स्थान / Space
D रंग / Colour
Explanation opens after your attempt
Correct Answer
B. आकार / Shape
Step 1
Concept
Circle square and triangle are geometric shapes. Exam tip: call closed flat figures shapes.
Step 2
Why this answer is correct
The correct answer is B. आकार / Shape. Circle square and triangle are geometric shapes. Exam tip: call closed flat figures shapes.
Step 3
Exam Tip
वृत्त वर्ग और त्रिभुज ज्यामितीय आकार हैं। परीक्षा में बंद समतल आकृतियों को आकार कहें।
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संख्या रेखा पर \( \sqrt{5} \) बनाने के लिए (2) और (1) लंबाई वाली समकोण भुजाओं का कर्ण क्या होगा?
To construct \( \sqrt{5} \) on the number line, what is the hypotenuse of a right triangle with perpendicular sides (2) and (1)?
#number-line
#geometric-construction
#pythagoras
A (3)
B \( \sqrt{3} \)
C \( \sqrt{5} \)
D (5)
Explanation opens after your attempt
Correct Answer
C. \( \sqrt{5} \)
Step 1
Concept
By Pythagoras, the hypotenuse is \( \sqrt{2^2+1^2}=\sqrt{5} \). Right triangles help in square-root construction.
Step 2
Why this answer is correct
The correct answer is C. \( \sqrt{5} \). By Pythagoras, the hypotenuse is \( \sqrt{2^2+1^2}=\sqrt{5} \). Right triangles help in square-root construction.
Step 3
Exam Tip
पायथागोरस से कर्ण \( \sqrt{2^2+1^2}=\sqrt{5} \) होगा। वर्गमूल निर्माण में समकोण त्रिभुज उपयोगी है।
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संख्या रेखा पर \(\sqrt{17}\) बनाने के लिए समकोण त्रिभुज की कौन-सी भुजाएँ सबसे उपयुक्त हैं?
Which legs of a right triangle are most suitable to construct \(\sqrt{17}\) on the number line?
#polynomials
#number-line
#square-root-construction
#pythagoras
A (4) और (1) / (4) and (1)
B (3) और (2) / (3) and (2)
C (5) और (1) / (5) and (1)
D (4) और (2) / (4) and (2)
Explanation opens after your attempt
Correct Answer
A. (4) और (1) / (4) and (1)
Step 1
Concept
Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 2
Why this answer is correct
The correct answer is A. (4) और (1) / (4) and (1). Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 3
Exam Tip
क्योंकि \(4^2+1^2=17\), इसलिए कर्ण \(\sqrt{17}\) होगा। संख्या रेखा पर वर्गमूल निर्माण में पाइथागोरस प्रमेय उपयोगी है।
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संख्या रेखा पर \(\sqrt{3}\) बनाने में समकोण त्रिभुज की कौन-सी भुजाएँ उपयोगी हो सकती हैं?
Which legs of a right triangle can be useful to construct \(\sqrt{3}\) on the number line?
#polynomials
#number-line
#construction
#pythagoras
A \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
B (2) और (1) / (2) and (1)
C (3) और (1) / (3) and (1)
D \(\sqrt{3}\) और (1) / \(\sqrt{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
Step 1
Concept
Because (\(\sqrt{2}\)2 +12 =3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1). Because (\(\sqrt{2}\)2 +12 =3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 3
Exam Tip
क्योंकि (\(\sqrt{2}\)2 +12 =3), अतः कर्ण \(\sqrt{3}\) होगा। क्रमिक वर्गमूल बनाने में पाइथागोरस का प्रयोग होता है।
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संख्या रेखा पर \(\sqrt{29}\) बनाने के लिए समकोण त्रिभुज की कौन सी दो लंब भुजाएं सही हो सकती हैं?
To construct \(\sqrt{29}\) on the number line, which two perpendicular sides of a right triangle can be correct?
#number-line
#construction
#pythagoras
#square-root
A (5) और (2) / (5) and (2)
B (4) और (4) / (4) and (4)
C (3) और (5) / (3) and (5)
D (6) और (1) / (6) and (1)
Explanation opens after your attempt
Correct Answer
A. (5) और (2) / (5) and (2)
Step 1
Concept
\(5^2+2^2=29\), so the hypotenuse will be \(\sqrt{29}\). Check the sum of squares of the sides.
Step 2
Why this answer is correct
The correct answer is A. (5) और (2) / (5) and (2). \(5^2+2^2=29\), so the hypotenuse will be \(\sqrt{29}\). Check the sum of squares of the sides.
Step 3
Exam Tip
\(5^2+2^2=29\), इसलिए कर्ण \(\sqrt{29}\) होगा। भुजाओं के वर्गों का योग जांचें।
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समकोण त्रिभुज में भुजाएं (3) और (4) लेकर संख्या रेखा पर कौन सा मान बनाया जा सकता है?
Using sides (3) and (4) in a right triangle, which value can be constructed on the number line?
#number-line
#construction
#pythagoras
#real-numbers
A (5)
B (6)
C (7)
D (12)
Explanation opens after your attempt
Step 1
Concept
The hypotenuse is \(\sqrt{3^2+4^2}=5\). This is a direct use of a Pythagorean triple.
Step 2
Why this answer is correct
The correct answer is A. (5). The hypotenuse is \(\sqrt{3^2+4^2}=5\). This is a direct use of a Pythagorean triple.
Step 3
Exam Tip
कर्ण \(\sqrt{3^2+4^2}=5\) होता है। यह पाइथागोरस त्रिक का सीधा उपयोग है।
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संख्या रेखा पर \(\sqrt{17}\) बनाने के लिए समकोण त्रिभुज की लंब भुजाएं (4) और (1) ली गई हैं। कर्ण की लंबाई क्या होगी?
To construct \(\sqrt{17}\) on the number line, the perpendicular sides of a right triangle are taken as (4) and (1). What will be the length of the hypotenuse?
#number-line
#square-root
#construction
#pythagoras
A \(\sqrt{15}\)
B \(\sqrt{17}\)
C (5)
D (17)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{17}\)
Step 1
Concept
The hypotenuse is \(\sqrt{4^2+1^2}=\sqrt{17}\). Use Pythagoras in this type of construction.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{17}\). The hypotenuse is \(\sqrt{4^2+1^2}=\sqrt{17}\). Use Pythagoras in this type of construction.
Step 3
Exam Tip
कर्ण \(=\sqrt{4^2+1^2}=\sqrt{17}\) होगा। ऐसी रचना में पाइथागोरस प्रमेय लगाएं।
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संख्या रेखा पर \(\sqrt{13}\) को बनाने के लिए (3) और (2) लंब भुजाओं वाला समकोण त्रिभुज बनाया जाए तो कर्ण क्या होगा?
To construct \(\sqrt{13}\) on the number line, if a right triangle has perpendicular sides (3) and (2), what is the hypotenuse?
#number-line
#construction
#pythagoras
#square-root
A \(\sqrt{5}\)
B \(\sqrt{11}\)
C \(\sqrt{13}\)
D (5)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{13}\)
Step 1
Concept
The hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Apply Pythagoras in a right triangle.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{13}\). The hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Apply Pythagoras in a right triangle.
Step 3
Exam Tip
कर्ण \(=\sqrt{3^2+2^2}=\sqrt{13}\) होता है। समकोण त्रिभुज में पाइथागोरस प्रमेय लागू करें।
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संख्या रेखा पर \(\sqrt{2}\) को बनाने के लिए समकोण त्रिभुज की दो लंब भुजाएं (1) और (1) ली गई हैं। कर्ण की लंबाई क्या होगी?
To construct \(\sqrt{2}\) on the number line, two perpendicular sides of a right triangle are taken as (1) and (1). What will be the length of the hypotenuse?
#number-line
#real-numbers
#square-root
#construction
A (1)
B \(\sqrt{2}\)
C (2)
D \(\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{2}\)
Step 1
Concept
By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 3
Exam Tip
पाइथागोरस से कर्ण \(=\sqrt{1^2+1^2}=\sqrt{2}\) होगा। ऐसी रचनाओं में वर्गों का योग जरूर देखें।
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संख्या रेखा पर \(\sqrt{2}\) को बनाने में किस समकोण त्रिभुज की कर्ण लंबाई उपयोगी है?
Which right triangle hypotenuse is useful to construct \(\sqrt{2}\) on the number line?
#sqrt-construction
#pythagoras
#number-line
A भुजाएँ (1) और (1) / Legs (1) and (1)
B भुजाएँ (2) और (2) / Legs (2) and (2)
C भुजाएँ (1) और (2) / Legs (1) and (2)
D भुजाएँ (3) और (1) / Legs (3) and (1)
Explanation opens after your attempt
Correct Answer
A. भुजाएँ (1) और (1) / Legs (1) and (1)
Step 1
Concept
If the legs are (1) and (1), the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This length is transferred to the number line with an arc.
Step 2
Why this answer is correct
The correct answer is A. भुजाएँ (1) और (1) / Legs (1) and (1). If the legs are (1) and (1), the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This length is transferred to the number line with an arc.
Step 3
Exam Tip
समकोण त्रिभुज की भुजाएँ (1) और (1) होने पर कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। इसी लंबाई को संख्या रेखा पर चाप से रखते हैं।
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एक समकोण त्रिभुज की भुजाएँ (x), (x+7) और कर्ण (x+8) हैं। छोटी भुजा क्या है?
The sides of a right triangle are (x), (x+7), and hypotenuse (x+8). What is the smallest side?
#quadratic equations
#pythagoras
#reasoning
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+7)2 =(x+8)2 ). Solving gives (x=5).
Step 2
Why this answer is correct
The correct answer is B. (5). By Pythagoras, (x-2 +(x+7)2 =(x+8)2 ). Solving gives (x=5).
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+7)2 =(x+8)2 )। हल करने पर (x=5) है।
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एक समकोण त्रिभुज में एक भुजा (10,सेमी) है और कर्ण दूसरी भुजा से (5,सेमी) अधिक है। दूसरी भुजा क्या है?
In a right triangle, one side is (10,cm) and the hypotenuse is (5,cm) more than the other side. What is the other side?
#quadratic equations
#pythagoras
#word problem
A (6.5,सेमी) / (6.5,cm)
B (7,सेमी) / (7,cm)
C (8,सेमी) / (8,cm)
D (7.5,सेमी) / (7.5,cm)
Explanation opens after your attempt
Correct Answer
D. (7.5,सेमी) / (7.5,cm)
Step 1
Concept
If the other side is (x), then (x-2 +102 =(x+5)2 ). This gives (x=7.5).
Step 2
Why this answer is correct
The correct answer is D. (7.5,सेमी) / (7.5,cm\(). If the other side is (x), then (x^2+10^2=(x+5)^2). This gives (x=7.5).\)
Step 3
Exam Tip
दूसरी भुजा (x) हो तो (x-2 +102 =(x+5)2 )। इससे (x=7.5) मिलता है।
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एक समकोण त्रिभुज का कर्ण (25 cm) है और एक भुजा दूसरी से (5 cm) अधिक है। छोटी भुजा क्या है?
The hypotenuse of a right triangle is (25 cm), and one side is (5 cm) more than the other. What is the shorter side?
#quadratic equations
#pythagoras
#geometry
A (10 cm)
B (15 cm)
C (20 cm)
D (24 cm)
Explanation opens after your attempt
Correct Answer
B. (15 cm)
Step 1
Concept
Let the shorter side be (x), then (x-2 +(x+5)2 =252 ). This gives \(x^2+5x-300=0\), so (x=15).
Step 2
Why this answer is correct
The correct answer is B. (15 cm\(). Let the shorter side be (x), then (x^2+(x+5)^2=25^2). This gives (x^2+5x-300=0), so (x=15).\)
Step 3
Exam Tip
छोटी भुजा (x) हो, तो (x-2 +(x+5)2 =252 )। इससे \(x^2+5x-300=0\), इसलिए (x=15)।
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एक समकोण त्रिभुज में लम्ब भुजाएँ (x) मीटर और (x+9) मीटर हैं तथा कर्ण (45) मीटर है। छोटी लम्ब भुजा क्या है?
In a right triangle, the legs are (x) m and (x+9) m, and the hypotenuse is (45) m. What is the smaller leg?
#quadratic equations
#pythagoras
#application
A (24) मीटर / (24) m
B (27) मीटर / (27) m
C (30) मीटर / (30) m
D (36) मीटर / (36) m
Explanation opens after your attempt
Correct Answer
B. (27) मीटर / (27) m
Step 1
Concept
(x-2 +(x+9)2 =452 ) gives (x=27). A negative root is not taken for length.
Step 2
Why this answer is correct
The correct answer is B. (27) मीटर / (27) m. (x-2 +(x+9)2 =452 ) gives (x=27). A negative root is not taken for length.
Step 3
Exam Tip
(x-2 +(x+9)2 =452 ) से (x=27) मिलता है। लंबाई के लिए ऋणात्मक हल नहीं लिया जाता।
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एक समकोण त्रिभुज में लम्ब भुजाएँ (x) मीटर और (x+4) मीटर हैं तथा कर्ण (20) मीटर है। (x) क्या है?
In a right triangle, the legs are (x) m and (x+4) m and the hypotenuse is (20) m. What is (x)?
#quadratic equations
#pythagoras
#application
A (8)
B (10)
C (12)
D (16)
Explanation opens after your attempt
Step 1
Concept
(x-2 +(x+4)2 =202 ) gives (x=12). For length, accept only the positive solution.
Step 2
Why this answer is correct
The correct answer is C. (12). (x-2 +(x+4)2 =202 ) gives (x=12). For length, accept only the positive solution.
Step 3
Exam Tip
(x-2 +(x+4)2 =202 ) से (x=12) मिलता है। लंबाई के लिए केवल धनात्मक हल स्वीकार करें।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+2) सेमी हैं तथा कर्ण (10) सेमी है। (x) का धनात्मक मान क्या है?
The legs of a right triangle are (x) cm and (x+2) cm, and the hypotenuse is (10) cm. What is the positive value of (x)?
#quadratic equations
#pythagoras
#word problem
A (4)
B (5)
C (6)
D (8)
Explanation opens after your attempt
Step 1
Concept
(x-2 +(x+2)2 =102 ) gives the positive solution (x=6). A negative root is not used for length.
Step 2
Why this answer is correct
The correct answer is C. (6). (x-2 +(x+2)2 =102 ) gives the positive solution (x=6). A negative root is not used for length.
Step 3
Exam Tip
(x-2 +(x+2)2 =102 ) से धनात्मक हल (x=6) है। लंबाई के लिए ऋणात्मक हल नहीं लिया जाता।
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एक समकोण त्रिभुज में छोटी भुजा (x) सेमी और दूसरी लम्ब भुजा (x+7) सेमी है। कर्ण (13) सेमी है। (x) क्या है?
In a right triangle, the smaller leg is (x) cm and the other leg is (x+7) cm. The hypotenuse is (13) cm. What is (x)?
#quadratic equations
#pythagoras
#application
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+7)2 =132 ), giving (x=5). In a right triangle, always take the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is C. (5). By Pythagoras, (x-2 +(x+7)2 =132 ), giving (x=5). In a right triangle, always take the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+7)2 =132 ), जिससे (x=5) है। समकोण त्रिभुज में हमेशा कर्ण को सबसे बड़ी भुजा लें।
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किस विकल्प में ज्यामितीय आकारों का समूह है?
Which option is a group of geometric shapes?
#geometric shapes
#square circle triangle
A पत्ती बादल धुआं / Leaf cloud smoke
B कपड़ा पानी लहर / Cloth water wave
C बाल फर शाखा / Hair fur branch
D वर्ग वृत्त त्रिभुज / Square circle triangle
Explanation opens after your attempt
Correct Answer
D. वर्ग वृत्त त्रिभुज / Square circle triangle
Step 1
Concept
Square circle and triangle are regular geometric shapes. Exam tip: remember measured shapes as geometric.
Step 2
Why this answer is correct
The correct answer is D. वर्ग वृत्त त्रिभुज / Square circle triangle. Square circle and triangle are regular geometric shapes. Exam tip: remember measured shapes as geometric.
Step 3
Exam Tip
वर्ग वृत्त और त्रिभुज नियमित ज्यामितीय आकार हैं। परीक्षा में मापित आकारों को ज्यामितीय याद रखें।
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तीन भुजाओं वाला आकार कौन सा है?
Which shape has three sides?
#triangle
#three sides
#shape
A त्रिभुज / Triangle
B वर्ग / Square
C वृत्त / Circle
D अंडाकार / Oval
Explanation opens after your attempt
Correct Answer
A. त्रिभुज / Triangle
Step 1
Concept
A triangle has three sides. Exam tip: connect three sides with triangle.
Step 2
Why this answer is correct
The correct answer is A. त्रिभुज / Triangle. A triangle has three sides. Exam tip: connect three sides with triangle.
Step 3
Exam Tip
त्रिभुज में तीन भुजाएं होती हैं। परीक्षा में three sides को triangle से जोड़ें।
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किस विकल्प में केवल द्वि आयामी आकार हैं?
Which option has only two-dimensional shapes?
#shape
#two dimensional
#circle triangle
A घन और गोला / Cube and sphere
B बेलन और शंकु / Cylinder and cone
C ईंट और गेंद / Brick and ball
D वृत्त और त्रिभुज / Circle and triangle
Explanation opens after your attempt
Correct Answer
D. वृत्त और त्रिभुज / Circle and triangle
Step 1
Concept
Circle and triangle are flat shapes. Exam tip: two-dimensional shapes do not have depth.
Step 2
Why this answer is correct
The correct answer is D. वृत्त और त्रिभुज / Circle and triangle. Circle and triangle are flat shapes. Exam tip: two-dimensional shapes do not have depth.
Step 3
Exam Tip
वृत्त और त्रिभुज समतल आकार हैं। परीक्षा में द्वि आयामी में गहराई नहीं होती।
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