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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

∴ BC = 4 cm

Using Pythagoras theorem,

AB

AB

⇒ AB

⇒

Ans.

The required ratio is 5: 11.

Ans.

Also, D is the midpoint of BC.

∴

In right ΔABC, we have

AC

⇒ AC

⇒ AC

In ΔABD, we have

AD

⇒ AB

Putting the value in (i), we have

AC

⇒ AC

Ans.

⇒ ∠ABC = ∠ACB (Equal sides have equal opposite angles)

Now, in ΔABD and ΔECF

∠ABD = ∠ECF (Proved above)

∠ADB = ∠EEC (Each 90°)

So, ΔABD ∼ ΔECF (By AA similarity)

Ans.

Given: Let, the equilateral triangle formed on one side of the square is ΔAED and another equilateral triangle formed on its diagonal BD is ΔBDF

Ans.

**Long Answer Type Question**

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

Ans.

AE

and BD

Adding (i) and (ii), we have

(∵ AC

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

Ans.

To Prove: ∠B = 90°.

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

Proof: In ΔPQR, we have,

PR

or, PR

But AC

So, AC

∴ AC = PR ...(iii)

Now, in ΔABC and ΔPQR,

AB = PQ (By construction)

BC = QR (By construction)

AC = PR (Proved in (iii))

So, ΔABC ≌ ΔPQR (By SSS congruency)

Therefore,

∠B = ∠Q (CPCT)

But ∠Q = 90° (By construction)

So, ∠B = 90°

Second part:

In ΔPQR,

By Pythagoras Theorem, we have

PR

⇒ PR

Now, in ΔPKR we have

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

Ans.

To prove: BC

Construction: Draw AD ⊥ BC

Proof: In ΔBDA and ΔBAC

∴ ΔBDA ~ ΔBAC (AA similarity)

⇒

(Corresponding sides of similar triangles)

⇒ AB

Also, in ΔADC and ΔBAC,

∠ADC = ∠BAC (Each 90°)

∠ACD = ∠ACB (Common)

∴ ΔADC ~ ΔBAC (AA similarity)

∴

(Corresponding sides of similar triangles)

⇒ AC

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

∴ AB

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