Q1. The length of the base of a triangle is 4 cm smaller than the length of its altitude. The area of the triangle is 96 cm^{2}. The length of the base is

A. 24 cm

B. 12 cm

C. 20 cm

D. 16 cm

E. 36 cm

Sol : Option B Explanation A = 96cm^{2}.
B – H = 4.
Also, 1/2 B.H = 96
∴ H (H – 4) = 192
⇒ H = 16 and B = 12 cm

Q2. The three sides of a triangle are 5m, 6m and 7m respectively, and then what is the area of the given triangle.

A. 15 m^{2}

B. 6&redic;6 m^{2}

C. 21 m^{2}

D. 7&redic;6 m^{2}

E. 32 m^{2}

Sol : Option B Explanation: Three sides of the triangle are 5 m, 6 m and 7 m. ∴s = (5 + 6 + 7)/2 = 9m. Putting the values, we get area =

Q3. The perimeter of an isosceles triangle is 100 cm. If the base is 36 cm, find the length of the equal sides.

A. 18 cm

B. 64 cm

C. 32 cm

D. 36 cm

E. 24 cm

Sol : Option C Explanation: Let the length of equal side = x.
∴ x + x + 36 = 100
⇒ 2x = 64 ⇒ x = 32cm.

Q4. A triangle has an area of 615 cm^{2}. One of its sides is given as 123 centimeters, and then what is the length of the perpendicular that is dropped on that particular side from the opposite vertex.

A. 10 cm

B. 5 cm

C. 0.12 cm

D. 0.2 cm

E. Data insufficient answer the question

Sol : Option A Explanation: A = 615cm^{2}. Base = 123 cm.
Perpendicular =?
Also A = 1/2 * B * H
∴ 615 = 1/2 * 123 * H
⇒ H = 10 cm.

5. Two sides of an isosceles triangle are 12.5 cm each while the third side is 20 cm. What is the area of the triangle?

A. 250 cm^{2}

B. 75 cm^{2}

C. 125 cm^{2}

D. 150 cm^{2}

E. None of these

Sol : Option B Explanation: Lengths of the sides are 12.5 cm, 12.5 cm and 20 cm. ∴ s = (12.5 + 12.5 + 20) / 2 ⇒ s = 22.5.

Q.6. In an isosceles right-angled triangle, the perimeter is 20 meters. Find its area.

A. 17.16 m^{2}

B. 5.858 m^{2}

C. 34.32 m^{2}

D. 9.320 m^{2}

E. 15.45 m^{2}

Sol : Option A Explanation: In an isosceles right angled triangle, Base = Height. Let a be the base and b be the hypotenuse.
∴ a + a + b = 20 ⇒ 2a + b = 20.
Also b^{2}= a^{2} + a^{2}
⇒ b^{2} = 2a^{2}.
∴ b =&redic;2a. So 2a+ &redic;2a = 20 ⇒ 3.41a = 20
∴ a = 5.86m.
Required area = 1/2 a^{2} = 1/2 * 5.86 * 5.86 =17.16 m^{2}

Q.7. Which of the following statements is true based on the following diagram?

A. KM < KL

B. KM < LM

C. LM < KL

D. KL < LM

E. None of these

Sol : Option C Explanation: Since ∠M> ∠K, KL>LM

Q8. In ∆LMN, XY ∥ MN, area of quadrilateral XMNY = 42 sq. m. If LX : XM = 2: 3, then find the area of ∆LXY.

A. 28 sq. m

B. 56/3 sq. m

C. 8 sq. m

D. 33.6 sq. m

E. 30.7 sq. m

Sol : Option C Explanation: XY ∥ MN
Therefore, ∆ LXY – ∆LMN …….. (AA Test)
Area (∆LXY) / Area (∆ LMN) = LX^{2} / LM^{2} (Let area (∆LXY = x))
∴ x / (x+Area (XMNY)) = (LX / LM)^{2}
∴ x / (x + 42) = (2/5) ^{2}
∴ x / (x + 42) = (4/25)
25x = 4x + 168
⇒ 21 x = 168
⇒ 8 sq.m

Q9. ∆ABC is a right-angled triangle. BD ⊥ AC. If and DC = 2 cm, then find the length of BD.

A. 4 cm

B. 4.5 cm

C. 5 cm

D. Data insufficient

Sol : Option A
∆ADB - ∆BDC
∴ AD/BD = BD/DC
∴ BD^{2} = AD * DC = 8 * 2
∴ BD^{2} = 16 ∴ BD = 4cm

Q10. In the following figure, the equilateral triangle ABC has an area of 900&redic;3m^{2}. Points P and Q are the mid points of the side AB and AC respectively. Find the area of the dotted region.

A. 64&redic;3m^{2}

B. 80&redic;3m^{2}

C. 75&redic;3m^{2}

D. 72&redic;3m^{2}

E. 60&redic;3m^{2}

Sol : Option C

Here side of the equilateral triangle is 60.In equilateral triangle PQ ∥ BC,
(∴ P & Q are mid points of the sides)
So BC = 2PQ ⇒ 2PQ = 30 m. Now AE=&redic;3/2 * 60 = 30&redic;3 m
As, O is a centroid. So AO : OE = 2 : 1
OE = 10&redic;3 m and AO = 20&redic;3 m
AF = 1/2 AE = 15&redic;3 m
FO = AO-AF=5&redic;3 m^{2}
Therefore area of ∆ PQO: 1/2 * PQ * FO
= 1/2 * 30 * 5&redic;3 = 75&redic;3 m^{2}