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**Short Answer Type Question **

**Q.1. In Fig., ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB. [CBSE, Allahabad 2019]Ans.** Given: AC = 4 cm

For isosceles triangle,

AC = BC

âˆ´ BC = 4 cm

Using Pythagoras theorem,

AB

AB

â‡’ AB

â‡’

Ans.

Required ratio is 5 : 11.

Ans.

Also, D is the mid point of BC.

âˆ´

In right Î”ABC, we have

AC

â‡’ AC

â‡’ AC

In Î”ABD, we have

AD

â‡’ AB

Putting the value in (i), we have

AC

â‡’ AC

Ans.

â‡’ âˆ ABC = âˆ ACB (Equal sides have equal opposite angles)

Now, in Î”ABD and Î”ECF

âˆ ABD = âˆ ECF (Proved above)

âˆ ADB = âˆ EEC (Each 90Â°)

So, Î”ABD âˆ¼ Î”ECF (By AA similarity)

Ans.

Given: Let, the equilateral triangle formed on one side of square is Î”AED and another equilateral triangle formed on its diagonal BD is Î”BDF

Ans.

**Long Answer Type Question**

**Q.1. D and E are points on the sides CA and CB respectively o f a triangle ABC right angled at C. Prove that AE ^{2} + BD^{2} = AB^{2} + DE^{2}. [CBSE 2019]** In right angled Î”ACE and Î”DCB, we have

Ans.

AE

and BD

Adding (i) and (ii), we have

(âˆµ AC

Use the above theorem to find the measure of âˆ PKR.

Ans.

To Prove: âˆ B = 90Â°.

Construction: We construct a Î”PQR right-angled at Q such that PQ = AB and QR = BC

Proof: In Î”PQR, we have,

PR

or, PR

But AC

So, AC

âˆ´ AC = PR ...(iii)

Now, in Î”ABC and Î”PQR,

AB = PQ (By construction)

BC = QR (By construction)

AC = PR (Proved in (iii))

So, Î”ABC â‰Œ Î”PQR (By SSS congruency)

Therefore,

âˆ B = âˆ Q (CPCT)

But âˆ Q = 90Â° (By construction)

So, âˆ B = 90

Second part:

In Î”PQR,

By Pythagoras Theorem, we have

PR

â‡’ PR

Now, in Î”PKR we have

Hence, âˆ PKR = 90Â° (By Converse of Pythagoras Theorem)

Ans.

To prove: BC

Construction: Draw AD âŠ¥ BC

Proof: In Î”BDA and Î”BAC

âˆ´ Î”BDA ~ Î”BAC (AA similarity)

â‡’

(Corresponding sides of similar triangles)

â‡’ AB

Also, in Î”ADC and Î”BAC,

âˆ ADC = âˆ BAC (Each 90Â°)

âˆ ACD = âˆ ACB (Common)

âˆ´ Î”ADC ~ Î”BAC (AA similarity)

âˆ´

(Corresponding sides of similar triangles)

â‡’ AC

On adding (i) and (ii), we get

âˆ´ AB

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