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# Geometry Word Problems

Geometry Word Problems
• Page 1
1.
In the figure, θ = $\pi$/2, = . Find the area of the circle in terms of $y$ where $x$ is the diameter of the circle.

 a. b. c. d.

#### Solution:

x is the diameter of the circle.
[θ = 90Ã‚Âº.]

x = 5y 6
[Given.]

Area of the circle = πd24
[Formula.]

= 25πy2 36
[Substitute the value of x in d and simplify.]

Correct answer : (2)
2.
In Δ PQR, A, B and C are points on PQ, QR and PR such that . If the area of ΔPQR = 100 sq.cm, then what is the area of ΔABC in sq.cm?

 a. 28 sq. cm b. 44.44 sq.cm c. 36 sq.cm d. 16.67 sq.cm

#### Solution:

PA / AQ =QB / BR =RC / CP =23
[Given.]

PA / PQ =25;PC / PR =35
[Step 1.]

QB / QR =25;QA / PQ =35
[Step 1.]

RC / RP =25;RB / RQ =35
[Step 1.]

Area of Δ PAC is proportional to the product of PA and PC.
[Area = half the product of two sides and sine of the included angle.]

Area of Δ PQR is proportional to the product of PQ and PR.
[Area = half the product of two sides and sine of the included angle.]

Area of Δ PAC = 25×35 of area of Δ PQR
[Area proportional to the product of the ratio of the sides.]

= 625 of area of Δ PQR
[Area proportional to the product of the ratio of the sides.]

Similarly Areas of Δ AQB and ΔBRC are also 625 of the area of Δ PQR

Area of Δ ABC = Area of Δ PQR - (Area of Δ PCA + Area of Δ ABQ + Area of Δ BCR)

= 100 - 3× 6×10025 = 28 sq.cm

Correct answer : (1)
3.
In the given figure, $\stackrel{‾}{\mathrm{AB}}$, $\stackrel{‾}{\mathrm{EF}}$ and $\stackrel{‾}{\mathrm{CD}}$ are parallel lines. EG = 3 cm, GC = 6 cm, DC = 8 cm and AB = 8 cm. Find $\stackrel{‾}{\mathrm{AC}}$.

 a. 12 cm b. 4.5 cm c. 17 cm d. 18 cm

#### Solution:

ΔEGF ~ ΔCGD

EF / DC = EG / CG
[Corresponding sides of similar triangles are proportional.]

EF / 8 = 3 / 6
[Substitute.]

EF = 4 cm
[Simplify.]

ΔACB ~ ΔECF
[AA similarity postulate.]

AC / EC = AB / EF
[Corresponding sides of similar triangles are Proportional.]

AC / 3+6 = 8 / 4
[Substitute.]

AC = 72 / 4 = 18 cm
[Simplify.]

Correct answer : (4)
4.
In ΔABC, $\stackrel{‾}{\mathrm{PQ}}$ intersects $\stackrel{‾}{\mathrm{AB}}$ at P and $\stackrel{‾}{\mathrm{AC}}$ at Q such that $\stackrel{‾}{\mathrm{PQ}}$ || $\stackrel{‾}{\mathrm{BC}}$ and divides ΔABC into two parts equal in area. Find $\frac{BP}{AB}$.

 a. 1 : 1 b. 2 : $\sqrt{2}$ c. ( d. 1 : $\sqrt{2}$

#### Solution:

In ΔAPQ and ΔABC, BAC = PAQ
[Common angle.]

APQ = ABC
[Corresponding angles.]

ΔAPQ ~ ΔABC
[AA similarity postulate.]

Area of ΔABC = Area of ΔAPQ + area of the trapezoid PQCB.

Area of ΔABC = 2(Area of ΔAPQ)
[Area of ΔAPQ = Area of the trapezoid PQCB.]

Area of ΔAPQ/ Area of ΔABC = 1 / 2 = AP2AB2
[ΔABC similar to ΔAPQ.]

AB2 = 2(AP2)
[Cross-product property.]

AB = 2(AP)
[Simplify.]

AB = 2(AB - BP)
[AP + PB = AB.]

(AB) (2 -1) = (BP)2
[Simplify.]

BP / AB = 2 - 12

Correct answer : (3)
5.
Find the value of $\frac{AD}{PK}$, if $y$ = 24, $x$ = 28, $r$ = 30 and $q$ = 35.

 a. $\frac{16}{25}$ b. $\frac{4}{5}$ c. 1 d. $\frac{6}{7}$

#### Solution:

ΔABC ~ ΔPQR
[SAS similarity theorem.]

BC / QR = AB / PQ
[ΔABC similar to ΔPQR.]

2(BD) / 2(QK) = AB / PQ
[BC = 2(BD) and QR = 2(QK).]

BD / QK = AB / PQ and BAD = QPK
[ΔABC similar to ΔPQR.]

ΔABD ~ ΔPQK
[SAS similarity theorem.]

AD / PK = AB / PQ
[ΔABD similar to Δ PQK.]

AD / PK = 24 / 30
[Substitute.]

AD / PK = 4 / 5
[Simplify.]

Correct answer : (2)
6.
In the ΔABC, $\stackrel{‾}{\mathrm{DE}}$ || $\stackrel{‾}{\mathrm{BC}}$. AD = 6 cm, DB = 9 cm, AE = 5 cm and BC = 13 cm. Find $\stackrel{‾}{\mathrm{DE}}$.

 a. $\frac{5}{26}$cm b. $\frac{26}{5}$cm c. 5 cm d. $\frac{13}{2}$cm

#### Solution:

DAE = BAC
[Same angle.]

ADE = ABC and AED = ACB
[Corresponding angles are congruent.]

ΔADE ~ ΔABC

AD / AB= DE / BC
[Corresponding sides of similar Δ′s are proportional.]

6 / 6+9 = DE / 13
[Substitute.]

DE = 6 / 15 × 13
[Cross multiply.]

DE = 26 / 5
[Simplify.]

Correct answer : (2)
7.
In the figure, $\stackrel{‾}{\mathrm{FG}}$ || $\stackrel{‾}{\mathrm{DE}}$ and $\stackrel{‾}{\mathrm{DE}}$ || $\stackrel{‾}{\mathrm{BC}}$. AE = 6 cm. Find the length of $\stackrel{‾}{\mathrm{EC}}$.

 a. 1.2 cm b. 4 cm c. 1.5 cm d. 2.4 cm

#### Solution:

AD / DB = AE / EC
[Side-splitter Theorem.]

3+5 / 2 = 6 / EC
[AD = AF + FD.]

EC = 2 × 6 / 8
[Cross-multiply.]

EC = 3 / 2 = 1.5 cm
[Simplify.]

Correct answer : (3)
8.
In the ΔABC, $\stackrel{‾}{\mathrm{DE}}$ || $\stackrel{‾}{\mathrm{BC}}$. AE = 10 cm, EC = 8 cm, YC = 2 cm and BY = 18 cm. Find $\stackrel{‾}{\mathrm{DX}}$.

 a. 12 cm b. 9 cm c. 10 cm d. 22 cm

#### Solution:

AX / XY = AE / EC
[Side-splitter theorem.]

AX / XY = 10 / 8
[Substitute.]

AX / AY = AE / AC = AE / AE+EC = 10 / 10+8= 10 / 18

DX / BY = AX / AY
[ΔADX similar to ΔABY.]

DX / 18 = 10 / 18
[Substitute.]

DX = 10 cm.

Correct answer : (3)
9.
In the figure, find the length of $\stackrel{‾}{\mathrm{ED}}$.

 a. 10 cm b. 4.2 cm c. 4 cm d. $5\frac{3}{5}$ cm

#### Solution:

EAD = CAB
[Same angle.]

AE / AB = 12 / 16+14 = 12 / 30 = 2 / 5

AD / AC = 16 / 12+28 = 16 / 40 = 2 / 5

AE / AB = AD / AC = 2 / 5
[Transitive property.]

ΔADE ~ ΔACB
[SAS similarity theorem.]

DE / CB = AD / AC
[Corresponding sides of similar triangles are proportional.]

DE / 14 = 2 / 5
[Step 4 & 6.]

ED = 16 × 14 / 40
[Cross-multiply.]

ED = 53 / 5 cm
[Simplify.]

Correct answer : (4)
10.
In the figure, find $\angle$BAC. [Given AB = 10 cm, BC = 5 cm, CD = 9 cm and DA = 18 cm.]

 a. $\angle$ADC b. $\angle$BCA c. $\angle$ABC d. $\angle$CAD

#### Solution:

BC / CD = 5 / 9

AB / AD= 10 / 18 = 5 / 9

BC / CD = AB / AD
[Transitive property.]

BAC CAD
[Converse of Triangle-Angle-Bisector Theorem.]

Correct answer : (4)

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