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Question

In the given figure, AD = AE, the measure of ∠DPB is

दिए गए चित्र में, AD = AE, तब ∠DPB का माप है

A

55°

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B

20°

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C

25°

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D

15°

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Solution

The correct option is D 15°

Solution

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Q. In the given figure, DE and BF are the bisectors of ∠ADC and ∠CBA respectively. If DE = BF, AD || BC, AD = BC, AE = 4 cm and BE = 6 cm, then FC + DF equals

दिए गए चित्र में, DE तथा BF क्रमशः ∠ADC तथा ∠CBA के समद्विभाजक हैं। यदि DE = BF, AD || BC, AD = BC, AE = 4 cm तथा BE = 6 cm है, तब FC + DF बराबर है

Q. In the given figure, ΔABD ≅ ΔCAD. If

A D = 3 \sqrt{2} c m,

then AB² – CD² (in cm²) equals

दिए गए चित्र में, ΔABD ≅ ΔCAD है। यदि

A D = 3 \sqrt{2} c m,

तब AB² – CD² (cm² में) बराबर है

Q. In the given figure ABC and ADE are triangles such that AB = AD, AE = EC . Find the measure of angle DAE

In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Q. This section contains 1 Assertion-Reason type question, which has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
इस खण्ड में 1 कथन-कारण प्रकार का प्रश्न है, जिसमें 4 विकल्प (a), (b), (c) तथा (d) दिये गये हैं, जिनमें से केवल एक सही है।

A : In the given figure, if ∠B = 60°, ∠C = 30°, AE is bisector of ∠A and AD ⊥ BC, then ∠DAE = 15°.
A : दिए गए चित्र में, यदि ∠B = 60°, ∠C = 30° और ∠A का समद्विभाजक AE है तथा AD ⊥ BC है, तब ∠DAE = 15°

R : In the above case,

∠ D A E = \frac{∠ C - ∠ B}{4}

R : उपरोक्त स्थिति में,

∠ D A E = \frac{∠ C - ∠ B}{4}

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