﻿ Similar Right Triangle Worksheet | Problems & Solutions

Similar Right Triangle Worksheet

Similar Right Triangle Worksheet
• Page 1
1.
Is any triangle among the triangles in ΔABC similar to ΔDBE?

 a. No b. Yes

Solution:

Similar figures have same shape but not necessarily same size.

ΔABC, ΔCAD, and ΔBCD are not similar to ΔPNS as they all are right triangles

ΔDEC is also not similar to ΔDBE
[From the figure.]

Among the triangles in ΔABC no triangle is similar to ΔDBE.

2.
ΔABC is similar to which of the following figures?

 a. Figure 1 only b. Figure 2 only c. Figure 3 only d. All the above

Solution:

Similar figures have same shape but not necessarily of same size.

ΔABC is a right triangle.
[From the figure.]

The three triangles represented by figure 1, figure 2, and figure 3 are right triangles.
[From the figure.]

As the ΔABC is a right triangle, the three triangles represented by the three figures have same shape as that of ΔABC and so they are similar to it.

3.
Solve for $y$. [Given $a$ = 7 and $b$ = 28.]

 a. 12 b. 14 c. 17 d. 7

Solution:

[BAC = CAD, ABC = ACD, AAA postulate, corresponding sides of similar triangles are proportional.]

28(y + 7) =(y + 7)y
[Substitute.]

(y + 7)2 = 28y
[Cross Multiply.]

y = 7
[Solve for y.]

4.
Solve for $v$. [Given $a$ = 2 and $b$ = 6.]

 a. 2 b. 6 c. 4 d. 5

Solution:

ΔRQS ~ ΔPQR RQ / QS = PQ / QR
[RQS = PQR, RSQ = PRQ, AAA postulate; corresponding sides of similar triangles are proportional.]

(v + 2)v  =(v + 6 )(v + 2)
[Substitute.]

(v + 2)2 = v(v + 6)
[Cross Multiply.]

v = 2
[Solve for v.]

5.
If ΔABC is a right triangle and $\stackrel{‾}{\mathrm{CD}}$ is altitude to the hypotenuse, then find $m$. [Given $a$ = 14 units, $b$ = 16 units.]

 a. 9.29 units b. 8.29 units c. 7.29 units d. 5.29 units

Solution:

ΔABC is a right triangle and CD is altitude to the hypotenuse
[Given.]

AD / CD = CD / DB
[Corresponding sides of similar triangles are proportional.]

2m = m14
[Substitute.]

m2 = 28 m = 5.29 units

6.
Find the perimeter of the right triangle ABC. [Given $a$ = 8, $b$ = 10.]

 a. 25.42 units b. 28.42 units c. 23.42 units d. 33.42 units

Solution:

ΔABC is a right triangle and CD is altitude to the hypotenuse.
[Given.]

[ADC = CDB, ACD = BCD, AAA postulate, corresponding sides of similar triangles are proportional.]

2m = m8
[Substitute.]

m2 = 16 m = 4 units

In ΔCDB, CB2 = DB2 + DC2
[Pythagorean theorem.]

CB2 = 82 + 42 CB = 8.94 units
[Substitute.]

In ΔABC, AB2 = CB2 + AC2
[Pythagorean theorem.]

102 = 8.942 + AC2 AC = 4.48 units
[Substitute and Simplify.]

Perimeter of ΔABC = AB + BC + AC = 10 + 8.94 + 4.48 = 23.42 units
[Substitute and Simplify.]

7.
Find the perimeter of the right triangle PQR. [Given $x$ = 8.6 and $y$ = 17.2.]

 a. 61.76 units b. 71.76 units c. 12.16 units d. 21.06 units

Solution:

ΔPQR is a right triangle and RS is altitude to the hypotenuse
[Given.]

ΔPQR ~ ΔPRS PQ / PR = PR / PS
[Angle PQR = PRS, PRQ = PSR, AAA postulate; corresponding sides of similar triangles are proportional.]

25.8a =a8.6 a2 = 221.88 a = 14.9 units
[Substitute and Simplify.]

ΔPRS ~ ΔQRS PS / RS = RS / SQ
[PRS = QRS, PSR = QSR, AAA postulate, corresponding sides of similar triangles are proportional.]

8.6b =b17.2
[Substitute.]

b2 = 147.92 b = 12.16 units
[Simplify.]

In ΔQRS, c2 = b2 + SQ2
[Pythagorean theorem.]

c2 = 12.162 + 17.22 c2 = 147.86 + 295.84 = 443.70
[Substitute.]

c = 21.06 units
[Simplify.]

The perimeter of ΔPQR is, PQ + PR + RQ = 25.8 + 21.06 + 14.9 = 61.76 units
[Substitute and Simplify.]

8.
If ΔPQR is a right triangle and $\stackrel{‾}{\mathrm{RS}}$ is altitude to the hypotenuse, then find the values of $r$ and $v$. [Given $a$ = 6.6 and $b$ = 11.]

 a. 10.8 units, 15.75 units b. 9.8 units, 13.75 units c. 8.8 units, 16.75 units d. 8.8 units, 13.75 units

Solution:

ΔPQR is a right triangle and RS is altitude to the hypotenuse
[Given.]

In ΔPSR, r2 = 112 - 6.62 r = 8.8 units
[Pythagorean theorem.]

ΔPSR ~ ΔRSQ
[PSR = RSQ, PRS = SQR, SPR = SRQ, AAA postulate.]

SP / SR = SR / SQ
[Corresponding sides of similar triangles are proportional.]

r6.6 =6.6SQ 8.8 / 6.6 = 6.6 / SQ
[Substitute.]

SQ = 43.56 / 8.8 units
[Simplify.]

v = SQ + r = 43.56 / 8.8 + 8.8
[Substitute.]

v = 121 / 8.8 = 13.75 units

9.
If ΔPQR is a right triangle and $\stackrel{‾}{\mathrm{RS}}$ is the altitude to the hypotenuse, then find the values of $a$, $b$, and $c$. [Given $x$ = 4.8 and $y$ = 9.6.]

 a. 11.75 units, 8.31 units, 6.78 units b. 8.31 units, 6.78 units, 11.75 units c. 10.31 units, 8.78 units, 13.75 units d. 6.78 units, 11.75 units, 8.31 units

Solution:

ΔPQR is a right triangle and RS is altitude to the hypotenuse
[Given.]

ΔPQR ~ ΔPRS PQ / PR= PR / PS
[PQR = PRS, PRQ = PSR, AAA postulate; corresponding sides of similar triangles are proportional.]

14.4a =a4.8 a2 = 69.12 a = 8.31 units
[Substitute and Simplify.]

ΔPRS ~ ΔRQS PS / RS = RS / SQ
[PRS = RQS, PSR = RSQ, AAA postulate; corresponding sides of similar triangles are proportional.]

4.8b = b9.6
[Substitute.]

b2 = 46.08 b = 6.78 units
[Simplify.]

In ΔQRS, c2 = b2 + SQ2
[Pythagorean theorem.]

c2 = 6.782 + 9.62 c2 = 45.96 + 92.16 = 138.12
[Substitute.]

c = 11.75 units
[Simplify.]

10.
Anny is at an Ice-cream parlor where, the path RP to her house and the path RQ to her school meet at right angle. She goes into the shopping complex, which is exactly opposite to the Ice-cream parlor. Find the distance between her house and the shopping complex. [Given $a$ = 174 and $b$ = 232.]

 a. 104.40 m b. 197.3 m c. 217.5 m d. 200.92 m

Solution:

In ΔPQR, PQ2 = RQ2 + PR2
[Pythagorean theorem.]

PQ2 = 1742 + 2322 = 84100 PQ = 290 m

ΔPQR ~ ΔPRS PQ / PR = PR / PS
[PQR = PRS, PRQ = PSR, AAA postulate; corresponding sides of similar triangles are proportional.]

290 / 174 = 174 / PS PS = 104.40 m
[Substitute and Simplify.]