Triangles - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option A

Since, ΔABC and ΔPQR are similar triangles.
then, ∠B = ∠Q = 83°
Thus, in ΔABC,
∠C = 180° - (∠A + ∠ B)
or, ∠C = 180° - (47° + 83°)
∠C = 50°

Answer: Option B

In Triangle ABD,
∠BAD + ∠B + 90° = 180°
or, ∠BAD + ∠B = 90° - - - - - - - (1)

Now, in Triangle ABC,
∠ACB + ∠B + ∠A = 180°
∠ACB + ∠B = 180° - 90°
∠ACB = 90° - ∠B - - - - - (2)
From (1) and (2), ∠BAD = ∠ACB
So, AC = CD

Answer: Option B

In δ PQR,
QR + 2 = 2PQ
QR = 2PQ - 2 - - - - - - - (1)
PR = PQ + 10 - - - - - - (2)
Perimeter = 40 m
PQ + QR + Rp = 40
Putting the value of PQ and QR from equation (1) and (2),
PQ + 2PQ - 2 + PQ + 10 = 40
4PQ = 32
PQ = 8 cm which is the smallest side of the triangle.

Answer: Option C

Since, AB and CD bisects each other at O,
Hence, BC = AD = 6 cm.

Answer: Option B

∠BAL + ∠B + 90° = 180°
or, ∠BAL + ∠B = 90°
or, ∠BAL = 90° - ∠B - - - - - - - - (1)
Now, in ΔABC,
∠ACB + ∠B + ∠A = 180°
∠ACB = 90°
-∠B - - - - - (2)
From, (1) and (2),
∠BAL = ∠ACB

Answer: Option A

Here, ∠ACB = c + 180 - (2c - b) = 180 - (b + c)
So, We can say that ΔBCD and ΔABC will be similar.
According to property of similarity,
$$\frac{{{\text{AB}}}}{{12}} = \frac{{12}}{9}$$
Hence,
AB = 16
$$\frac{{{\text{AC}}}}{6} = \frac{{12}}{9}$$
AC = 8
Hence, AD = 7 and AC = 8
Now,
$$\eqalign{ & \frac{{{\text{Perimeter of Delta ADC}}}}{{{\text{Perimeter of Delta BDC}}}} \cr & = \frac{{6 + 7 + 8}}{{9 + 6 + 12}} \cr & = \frac{{21}}{{27}} \cr & = \frac{7}{9} \cr} $$

Answer: Option B

Let BC = x and Ad = y, then as per bisector theorem,
$$\frac{{{\text{BD}}}}{{{\text{DC}}}} = \frac{{{\text{AB}}}}{{{\text{AC}}}} = \frac{4}{3}$$
Hence, BD = $$\frac{{4{\text{x}}}}{7}$$ and DC = $$\frac{{3{\text{x}}}}{7}$$
Now, in ΔABD using cosine rule,
$$\cos {30^ \circ } = \frac{{{4^2} + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$
$${\text{or,}}\,4\sqrt {3{\text{y}}} = {16 + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} } $$       - - - - - - (i)
Similarly in ΔADC,
$$\cos {30^ \circ } = \frac{{{3^2} + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$
$${\text{or,}}\,3\sqrt {3{\text{y}}} = {9 + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} } $$       - - - - - - - - (ii)
From equation (i) and (ii), we get
$${\text{y}} = \frac{{12\sqrt 3 }}{7}$$

Answer: Option C

Orthocentre.

Answer: Option D

∠1 + ∠2 = ∠3 [PS is bisector.] - - - - - - (1)
∠Q = 90° - ∠1
∠R = 90° -∠2 - ∠3
So,
∠Q - ∠R = (90° - ∠1) - (90° - ∠2 - ∠3)
∠Q - ∠R = ∠2 + ∠3 - ∠1
∠Q - ∠R = ∠2 + (∠1 + ∠2) -∠1[using equation 1]
∠Q - ∠R = 2∠2
$$\frac{1}{2}$$ × (∠Q - ∠R) = ∠TPS

Answer: Option A

Two right angled triangle are congruent if the hypotenuse of one triangle is equal to the hypotenuse of the other and a side of one triangle is equal to the corresponding side of the other triangle.