Question

ABCD is a rhombus with side equal to 8 cm. CB is produced to R ,such that BR = 4 cm. DR cuts AC and AB at P and Q respectively.
Find

ar(∆DCP)/ar(∆APQ)

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Answer 1

Answer:

ABCD is rhombus.

AC and BD are two diagonals of the ABCD rhombus.

All the angles, AOB, BOC, COD, AOD are 90°. (Rhombus's property)

Now,

For the right angle triangle BOA,

OB² + OA² = AB²

And, (OB = 1/2× BD = 10/2 = 5 cm)

Mathematical form will be,

(5)²+ (OA)² = (8)²

25 + (OA)2 = 64

(OA)2 = 64 - 25

OA = √39 cm

Now, AC = 2 × OA = √39 × 2 = 2√39 cm

As mentioned in the question,

⇒ 2√39 = 2√x

⇒ √x = √39

⇒ x = 39

So, the value of x + 1 is = 39 + 1 =40

Step-by-step explanation:

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