﻿ Triangle Word Problems | Problems & Solutions # Triangle Word Problems

Triangle Word Problems
• Page 1
1.
A, B and C are the mid points of the three sides of ΔPQR as shown in the figure. QB and CR intersect at G. If QD = 5 cm, find the length of GD.  a. $\frac{5}{3}$units b. 3 units c. 2 units d. $\frac{8}{5}$units

#### Solution:

ΔPQR ~ ΔCQA
[A and C are midpoints of QR and PQ.]

QB = 10
[QD = 5, Ratio of the sides of the similar triangles is 2.]

QG = 10 × 2 / 3 = 20 / 3
[G is the point of intersection of the medians, which divides the median in 2 : 1 ratio.]

GD = QG - QD = 20 / 3- 5 = 5 / 3units

2.
$\angle$DAC = 40o, find $\angle$FBC.  a. 25o b. 20o c. 50o d. 40o

#### Solution:

In ΔADC, ACD = 180o - (DAC + CDA)
[From the figure.]

180o - (40o + 90o) = 50o
[Substitute and simplify.]

In ΔBFC, FBC = 180o - (BCF + CFB)
[From the figure.]

FBC = 180o - (50o + 90o) = 40o
[Substitute and simplify.]

3.
If AD = 4 cm, CF = 3 cm, then what is the length of AC?  a. 2$\sqrt{5}$ cm b. 5$\sqrt{7}$ cm c. 5 cm d. 6$\sqrt{3}$ cm

#### Solution:

AB2 + AC2 = 2(AD2 + DC2) and AC2 + BC2 = 2(CF2 + FB2)
[Applying Apollonian theorem.]

(AB2 + BC2) + 2(AC2) = 2(AD2 + CF2) + 2(DC2 + FB2)

(AB2 + BC2) + 2(AC2) = 2[42 + 32] + 2(DC2 + FB2)

AC2 + 2AC2 = 2 x 25 + 2[(BC / 2)2 + (AB / 2)2]
[From the figure, DC = BC / 2, FB = AB / 2 and AB2 + BC2 = AC2.]

3AC2 = 50 + BC2 +AB22

3AC2 = 50 + AC22 AC2 = 20
[Simplify.]

AC = 25 cm
[Take the square root.]

4.
What is the point of concurrency of the medians of a triangle called? a. circumcenter b. orthocenter c. incenter d. centroid

#### Solution:

Centroid is the point of concurrency of the medians of the triangle.
[By definition.]

5.
Name the point of concurrency of the altitudes of a triangle. a. circumcenter b. orthocenter c. centroid d. incenter

#### Solution:

Orthocenter is the point of concurrency of the altitudes of the triangle.
[By definition.]

6.
In ΔABC, AD is the median and the area of the ΔABC is 36 cm2. Find the area of ΔADC.  a. 18 cm2 b. 6 cm2 c. 12 cm2 d. 16 cm2

#### Solution:

Median divides a triangle into two triangles of equal area.

Area of ΔADC = Area of ΔADB = 1 / 2 (Area of ΔABC)
[Step 1.]

Area of ΔADB = 1 / 2(36) = 18 cm2.
[Substitute.]

7.
In ΔABC, what is the length of the median AD?  a. 7$\sqrt{2}$ cm b. 18 cm c. 20 cm d. 24 cm

#### Solution:

AB2 + AC2 = 2(AD2 + DC2)
[By Apollonius Theorem.]

(152)2 + (202) 2 = 2[AD2 + 72]
[Substitute the values.]

450 + 800 = 2 [AD2] + 98
[Simplify]

[Simplify.]

[Divide by 2 on both sides.]

[Take square root on both sides.]

8.
In ΔABC, AD is the median and G is the Centroid. AG : AD = ?  a. 1 : 3 b. 2 : 3 c. 1 : 4 d. 1 : 2

#### Solution:

Centroid divides median in the ratio 2 : 1.

AG : GD = 2 : 1
[Step 1.]

AG : AD = 2 : 3
[Since AD = AG + GD = 2GD + GD = 3GD.]

9.
In what type of triangle at least one median coincides with an altitude? a. either isosceles or equilateral b. isosceles c. equilateral d. right angled

#### Solution: In an isosceles triangle, the median from the vertex containing the congruent sides is perpendicular to the base.

In an equilateral triangle, the median from any vertex to the opposite side will be perpendicular to that side.

So, in isosceles triangle and equilateral triangle, at least one median coincides with an altitude.

10.
Select the correct statement/statements.
1. In an equilateral triangle, orthocenter coincides with incenter.
2. In an equilateral triangle, median and angle bisector from a vertex are the same.
3. In an equilateral triangle, centroid coincides with circumcenter. a. 2 only b. 3 only c. 1, 2, and 3 d. 1 only

#### Solution:

In an equilateral triangle, median, altitude, perpendicular bisector and angle bisectors are same.

So, centroid, orthocenter, incenter and circumcenter of an equilateral triangle coincide.

All the statements are correct.