BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

In the figure below, what is x?

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

In the figure below, what is x?

by Max@Math Revolution » Wed Jan 22, 2020 5:15 pm
[GMAT math practice question]

In the figure below, what is x?

A. 3
B. 4
C. 5
D. 6
E. 7
1.22ps.png
Top
Source: — Problem Solving |
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

Re: In the figure below, what is x?

by Max@Math Revolution » Mon Jan 27, 2020 6:19 pm
=>

Since ∠CDA = 70 (from 180 - 110) and ∠CAD = 70, triangle ACD is an isosceles triangle and we have AC = CD.
Since ∠CAB = 110 (from 180 - 70) and ∠ABC = 35, we have ∠ACB = 35, triangle ABC is an isosceles triangle and we have AB = AC.
Thus we have AB = AC = CD.
Then we have x = 3 since AB = 3.

Therefore, A is the answer.
Answer: A
Top
Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pm
Followed by:5 members

Re: In the figure below, what is x?

by deloitte247 » Sat Feb 01, 2020 4:32 am
$$From\ \triangle ABC,\ length\ AB=3,\ \angle B=35^0$$
$$\triangle A=180^0-70^0\ =110^0\left(180^0=angle\ on\ a\ straight\ line\right)$$
$$\triangle C=180^0-\left(A^0+B^0\right)$$
$$\triangle C=180^0-\left(110^0+35^0\right)=35^0$$
Triangle ABC is an isosceles triangle with 2 equal angles (definitely 2 of its sides must also be equal)
Let AB=c, AC=b and BC=a
Using the sine rule
$$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$$
$$U\sin g;\ \frac{b}{SinB}=\frac{c}{SinC}=>\frac{b}{Sin35}=\frac{3}{Sin35}$$
$$b=3$$
From triangle ADC, length AC=3, Angle A=70 degrees
Angle D = 180 - 110 = 70
Angle C = 180 - (70+70) = 40
Triangle ADC is an isosceles triangle with 2 equal angles
Let AC=d, CD=x and AD=c
Using the sine rule;
$$\frac{d}{SinD}=\frac{x}{SinA}=\frac{c}{SinC}$$
$$U\sin g;\ \frac{d}{SinD}=\frac{x}{SinA}=>\frac{3}{Sin70}=\frac{x}{Sin70}$$
$$x=3$$

Answer = A
Top
Post new topic Post Reply
• Page 1 of 1