Triangles Test

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Triangles Test - 28

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Question 1
1 / -0

In triangle ABC, $$BC^2$$ + $$AB^2$$ = $$AC^2$$, then __________ is a right angle

A
$$\angle A$$

B
$$\angle B$$

C
$$\angle C$$

D
None

Solution

In $$\triangle ABC,$$

$$\Rightarrow$$ $$(AC)^2=(BC)^2+(AB)^2$$ [ Given ] --- ( 1 )

Pythagoras theorem,

$$\Rightarrow$$ $$(Hypotenuse)^2=(one\,side)^2+(other\,side)^2$$ ---- ( 2 )

Comparing ( 1 ) and ( 2 ) we get, $$AC$$ is a hypotenuse of a triangle.

$$\Rightarrow$$ Hypotenuse is the longest side of a right-angled triangle, opposite the right angle.

$$\Rightarrow$$ Opposite angle of $$AC$$ is $$\angle B.$$

$$\therefore$$ $$\angle B$$ is a right angle of $$\triangle ABC.$$
Question 2
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Two angles of triangle ABC are $$\displaystyle 85^{\circ}$$ and $$\displaystyle 65^{\circ}$$ while the two angles of another triangle DEF are $$\displaystyle 30^{\circ}$$ and $$\displaystyle 65^{\circ}$$.Which of the statements is correct?

A
$$\triangle$$ ABC is similar to $$\triangle$$ DEF

B
$$\triangle$$ ABC is congurent to $$\triangle$$ DEF

C
$$\triangle$$ ABC is equal to $$\triangle$$ DEF

D
None of these

Solution

For the first triangle, with two angles as 85$$^o$$ and 65$$^o$$ , let the third angle be $$x^o$$
Sum of angles = 180$$^o$$
$$85^o + 65 ^o + x^o = 180^o$$
$$x = 30^o$$
For the second triangle, with two angles as 30$$^o$$and 65$$^o$$, let the third angle be $$y$$
Sum of angles = $$180^o$$
$$85^o + 30^o + y^o = 180^o$$
$$y = 65^o$$
Since, the angles of the two triangles are equal. Hence, the two triangles are similar. (AAA rule)
Question 3
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The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If $$AE=4 cm, AF= 8$$ cm and AB $$=12$$ cm. find the perimeter of the parallelogram ABCD.

A
$$42$$ cm

B
$$44$$ cm

C
$$46$$ cm

D
$$48$$ cm

E
$$40$$ cm

Solution

In $$\triangle AFE$$ and $$\triangle BFC$$,
$$\angle AFE = \angle BFC$$ (Common angle)
$$\angle AEF = \angle BCF$$ (Corresponding angles of parallel lines)
$$\angle FAE = \angle CBF$$ (Corresponding angles of parallel lines)
Thus, $$\triangle AFE \sim \triangle CFB$$ ($$AAA$$ rule)
Hence, $$\dfrac{AF}{BF} = \dfrac{AE}{BC}$$
$$\dfrac{AF}{AF + AB} = \dfrac{AE}{BC}$$
$$\dfrac{8}{8 + 12} = \dfrac{4}{BC}$$
$$BC = \dfrac{4 \times 20}{8}$$
$$BC = 10$$ cm
Perimeter of parallelogram = $$AB + BC + BC+ AD$$
Perimeter of parallelogram = $$2 (AB + BC)$$
Perimeter of parallelogram = $$2 (12 + 10)$$ (Opposite sides are equal)
Perimeter of parallelogram = $$44$$ cm
Question 4
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$$\angle BAC$$ of triangle $$ABC$$ is obtuse and $$AB=AC$$. $$P$$ is a point in $$BC$$ such that $$PC= 12$$ cm. $$ PQ $$ and $$PR$$ are perpendiculars to sides $$AB$$ and $$AC$$ respectively. If $$PQ= 15$$ cm and $$=9$$ cm; find the length of $$PB$$.

A
$$20$$

B
$$24$$

C
$$36$$

D
$$18$$

Solution

Given: $$AB = AC$$, $$PQ \perp AB$$ and $$PR \perp AC$$
Since, $$AB = AC$$
$$\angle ABC = \angle ACB$$...(I) (Isosceles triangle property)

Now, In $$\triangle PBQ$$ and $$\triangle PRC$$
$$\angle PBQ = \angle PCR$$ (From I)
$$\angle PQB = \angle PRC$$ (Each $$90^{\circ}$$)
$$\angle QPB = \angle RPC$$ (Third angle)

Thus, $$\triangle QPB \sim \triangle RPC$$ (by AAA similarity)

If two triangles are similar, then their corresponding sides are proportional.
$$\dfrac{PQ}{PR} = \dfrac{PB}{PC}$$
$$\dfrac{15}{9} = \dfrac{PB}{12}$$
$$PB = \dfrac{15 \times 12}{9}$$
$$PB = 20\ cm$$
Question 5
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In the following figure, M is midpoint of BC of a parallelogram ABCD.
DM intersects the diagonal AC at P and AB produced at E. Then

A
$$PE = 3 PD$$

B
$$PE = 2 PD$$

C
$$PE = PD$$

D
$$PE = 4 PD$$

Solution

Given: ABCD is a parallelogram. M is mid point of BC.

In $$\triangle DMC$$ and $$\triangle MBE$$
$$\angle DMC = \angle BME$$ (Vertically opposite angles)
$$\angle DCM = \angle MBE$$ (Alternate angles)
$$MC = MB$$ (M is mid point of BC)
Thus, $$\triangle DMC \cong \triangle EMB$$ (ASA rule)
Thus, $$DC = BE$$ (By CPCT)

Now, In $$\triangle DPC$$ and $$\triangle APE$$
$$\angle DPC = \angle APE$$ (Vertically Opposite angles)
$$\angle CDP = \angle AEP$$ (Alternate angles)
$$\angle PCD = \angle PAE$$ (Alternate angles)
Thus, $$\triangle DPC \sim \triangle EPA$$ (AAA rule)
Hence, $$\frac{PE}{PD} = \frac{AE}{CD}$$
$$\frac{PE}{PD} = \frac{AB + BE}{CD}$$ (since AB = BE = CD)
$$\frac{PE}{PD} = 2$$
$$PE = 2 PD$$
Question 6
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In the figure below AB, CD and EF are parallel lines. Given AB$$=$$ 7.5 cm, DC$$=$$ y cm. EF $$=$$ 4.5 cm, BC$$=$$ x cm and CE $$=$$ 3 cm. Calculate the values of x and y.

A
x$$=$$ 1 cm and $$\displaystyle y=2\frac{13}{16}$$ cm

B
x$$=$$ 5 cm and $$\displaystyle y=2\frac{13}{16}$$ cm

C
x$$=$$ 7 cm and $$\displaystyle y=2\frac{13}{16}$$ cm

D
x$$=$$ 2 cm and $$\displaystyle y=2\frac{13}{16}$$ cm

Solution

In $$\triangle BCD$$ and $$\triangle BEF$$,
$$\angle DBC = \angle FBE$$ (Common angle)
$$\angle BCD = \angle BEF$$ (Corresponding angles of parallel lines CD and EF)
$$\angle BDC = \angle BFE$$ (Corresponding angles of parallel lines CD and EF)
Thus, $$\triangle BCD \sim \triangle BEF$$ (AAA rule)
Hence, $$\frac{BC}{BE} = \frac{BD}{BF} = \frac{CD}{EF}$$ (Corresponding sides)
$$\frac{x}{x + 3} = \frac{BD}{BF} = \frac{y}{4.5}$$ ..(I)

Similarly, $$\triangle FCD \sim \triangle FAB$$
Thus, $$\frac{FC}{FA} = \frac{CD}{AB} = \frac{FD}{FB}$$
$$\frac{y}{7.5} = \frac{DF}{BF}$$ (II)
From I and II,
$$BD = \frac{y}{4.5} BF$$ and $$DF = \frac{y}{7.5} BF$$
$$BD + DF = BF$$
$$\frac{y}{4.5}BF + \frac{y}{7.5} BF = BF$$
$$7.5 y + 4.5 y = 7.5 \times 4.5$$
$$y = \frac{45}{16}$$

Also, $$\frac{x}{x + 3} = \frac{y}{4.5}$$
$$\frac{x}{x + 3} = \frac{45}{16 \times 4.5}$$
$$\frac{x}{x + 3} = \frac{45}{72}$$
$$72x = 45x + 135$$
$$27x = 135$$
$$x = 5 cm $$
Question 7
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From the given figure, if ∠ABP=∠ACB, find AC

A
$$7.2$$ cm

B
$$12$$ cm

C
$$12.5$$ cm

D
$$11.25$$ cm

Solution

In $$\triangle ABC$$,
Given, $$\angle ABP = \angle ACB$$, $$AB= 9$$, $$BC = 15$$, $$BP = 12$$

In $$\triangle ABP$$ and $$\triangle ABC$$,
$$\angle ABP = \angle ACB$$ (Given)
$$\angle BAP = \angle BAC$$ (Common angle)
$$\angle APB = \angle ABC$$ (third angle)
Thus, $$\triangle ABP \sim \triangle ACB$$ (AAA rule)
Thus, $$\frac{AB}{AC} = \frac{BP}{BC}$$
$$\frac{9}{AC} = \frac{12}{15}$$
$$AC = \frac{9 \times 15}{12}$$
$$AC = 11.25$$ cm
Question 8
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A line segment $$DE$$ is drawn parallel to base $$BC$$ of $$\displaystyle \Delta ABC$$ which cuts $$AB$$ at point
$$D$$ and $$AC$$ at point $$E$$. If $$AB=5 BD$$ and $$EC = 3.2$$ cm. find the length of $$AE$$.

A
$$12.8$$ cm

B
$$11.8$$ cm

C
$$10.9$$ cm

D
none of the above
Question 9
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In the figure given below, $$AB /\parallel EF \parallel CD$$. If $$AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm$$ and $$DC = 27$$ cm. Find $$AC$$

A
$$AC = 37.5 $$ cm

B
$$AC = 22.5$$ cm

C
$$AC = 13.5$$ cm

D
$$AC = 27$$ cm

Solution

Given: $$AB \parallel EF \parallel CD$$
$$AB = 22.5, EP = 7.5, PC = 15, DC = 27$$

In $$\triangle DCP$$ and $$\triangle EPF$$,
$$\angle DCP = \angle PEF$$ (Alternate angles)
$$\angle CDP = \angle PFE$$ (Alternate angles)
$$\angle DPC = \angle EPF$$ (Vertically opposite angles)
Thus, $$\triangle DCP \sim \triangle FEP$$ (AAA rule)
Now, $$\dfrac{DC}{EF} = \dfrac{PC}{EP}$$
$$\dfrac{27}{EF} = \dfrac{15}{7.5}$$
$$EF = 13.5$$ cm

Now, In $$\triangle ABC$$ and $$\triangle ECF$$,
$$\angle CEF = \angle CAB$$ (Corresponding angles)
$$\angle CFE = \angle CBA$$ (Corresponding angles)
$$\angle ECF = \angle ACB$$ (Common angle)
Thus, $$\triangle ABC \sim \triangle EFC$$ (AAA rule)

Hence, $$\dfrac{AB}{EF} = \dfrac{AC}{EC}$$
$$\dfrac{22.5}{13.5} = \dfrac{AC}{7.5 + 15}$$
$$AC = \dfrac{22.5\times 22.5}{13.5}$$
$$AC = 37.5$$ cm
Question 10
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In the following figure, $$XY$$ is parallel to $$BC$$, $$AX= 9 $$ cm, $$XB= 4.5$$ cm and $$BC =18$$ cm. Find $$\displaystyle \dfrac{YC}{AC}$$

A
$$\displaystyle \dfrac{1}{3}$$

B
$$\displaystyle \dfrac{2}{3}$$

C
$$\displaystyle \dfrac{1}{2}$$

D
none of the above

Solution

In $$\triangle AXY$$ and $$\triangle ABC$$
$$\angle XAY = \angle BAC$$ (Common angle)
$$\angle AXY = \angle ABC$$ (Corresponding angles for parallel lines $$XY$$ and $$BC$$)
$$\angle AYX = \angle ACB$$ (Corresponding angles for parallel lines $$XY$$ and $$BC$$)
Thus, $$\triangle AXY \sim \triangle ABC$$ (AAA rule)

Therefore, $$ \dfrac { BX }{ AB } =\dfrac { YC }{ AC } $$
$$ \dfrac { YC }{ AC } =\dfrac { 4.5 }{ 4.5+9 } \\ \dfrac { YC }{ AC } =\dfrac { 4.5 }{ 13.5 } =\dfrac { 1 }{ 3 } $$