Triangles Test - 15

In ΔABC and ΔADC,

BC = CD = 2 cm (Given)

∠ABC = ∠ADC = 90° (Given)

AC = AC (Common)

∴ΔABC ≅ ΔADC [RHS congruence rule]

So, AB = AD = 5 cm [CPCT]

Thus, perimeter of quadrilateral ABCD = AB + BC + CD + AD

= (5 + 2 + 2 + 5) cm

= 14 cm

Hence, the correct answer is option B.

ΔABD is an isosceles triangle with AB = AD. It is known that angles opposite to equal sides are equal.

∴ ∠ABD = ∠ADB

Applying angle sum property in ΔABD:

∠ABD + ∠ADB + ∠BAD = 180°

⇒ 2∠ABD = 180° − 70° = 110°

⇒ ∠ABD = 55°

∴ ∠ABD = ∠ADB = 55°

Now, ∠ADB + ∠ADC = 180° (Linear pair)

⇒ ∠ADC = 180° − 55° = 125°

∠BAC = 90°

∴ ∠DAC = ∠BAC − ∠BAD = 90° − 70° = 20°

Applying angle sum property of triangles in ΔADC:

∠ADC + ∠ACD + ∠DAC = 180°

⇒ 125° + ∠ACD + 20° = 180°

⇒ ∠ACD = 180° − 145° = 35°

It is known that if two sides of a triangle are unequal, then the angle opposite to the longer side is larger.

∠DAC < ∠ACD

∴ CD< AD

Thus, the relation given in alternative B is not correct.

The correct answer is B.

It is given that BL = CM.

⇒ BL + LM = CM + LM

⇒ BM = CL

Consider ΔABM and ΔDCL.

∠ABM = ∠DCL = 90° (Given)

AM = DL (Given)

BM = CL (Shown above)

Therefore, by RHS congruency rule, ΔABM ≅ ΔDCL

∴ ∠BAM = ∠CDL (Corresponding angles of congruent triangles are equal)

∴ ∠CDL = 56°

Thus, the measure of ∠CDL is 56°.

The correct answer is A.

In ΔABC and ΔBAD,

BC = AD [Given]

AB = AB [Common]

AC = BD [Given]

∴ΔABC ≅ ΔBAD [By SSS congruency rule]

∴ ∠ABC = ∠DAB [By C.P.C.T.]

∴ OA = OB [Sides opposite to equal angles are equal]

BC = AD [Given]

⇒ AD = BC = 9 cm

OA = AD − OD = 9 cm − 7 cm = 2 cm

⇒ OB = OA = 2 cm

∴ Perimeter of ΔAOB = AB + OA + OB = (6 + 2 + 2) cm = 10 cm

Thus, the perimeter of ΔAOB is 10 cm.

The correct answer is A.