Triangles Test

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Triangles Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

In triangle ABC, AB = AC = 8 cm, BC = 4 cm and P is a point on AC, such that AP = 6 cm, then find the length of BP.

A
BP = 4 cm

B
BP = 8 cm

C
BP = 6 cm

D
BP = 12 cm

Solution

Given that, in $$\triangle ABC$$, $$AB = AC = 8\ cm$$, $$BC = 4\ cm$$ and $$AP = 6\ cm$$

In $$\triangle\,ABC$$,
$$\displaystyle\,\frac{AB}{BC}\,=\,\frac{8}{4}\,=\,2\quad ....(i)$$

$$AC=8\ cm\text{ and }AP= 6\ cm$$
$$\therefore \ CP=8-6=2\ cm$$

In $$\Delta\,BPC$$,
$$\displaystyle\,\frac{BC}{CP}\,=\,\frac{4}{2}\,=\,2\quad ....(ii)$$

From $$(i)$$ and $$(ii)$$, we get:
$$\displaystyle\,\frac{AB}{BC}\,= \displaystyle\,\frac{BC}{CP}=2$$

Also, angle $$C$$ is common in both triangles
$$\therefore \ \angle\,ACB\,=\,\angle\,PCB$$

Therefore, by SAS similarity criterion, $$\Delta\,ABC \sim \Delta\,BPC$$

Hence, $$\dfrac{AB}{BP} = \dfrac{AC}{BC}$$

$$\dfrac{8}{BP} = \dfrac{8}{4}$$

$$\therefore \ BP = 4\ cm$$
Question 2
1 / -0

In the figure given below, $$CD \parallel EF \parallel AB$$. If $$AB= 22.5$$ cm, $$EP= 7.5$$ cm, $$PC= 15$$ cm and $$DC=27$$ cm. Calculate $$EF$$.

A
$$EF= 14.5\ cm$$

B
$$EF= 13.5\ cm$$

C
$$EF= 12.5\ cm$$

D
None of the above

Solution

Given: $$CD \parallel EF \parallel AB$$

In $$\triangle DCP$$ and $$\triangle PEF$$,
$$\angle DPC = \angle EPF$$ (Vertically opposite angles)
$$\angle DCP = \angle PEF$$ (Alternate angles for parallel lines CD and EF)
$$\angle CDP = \angle PFE$$ (Alternate angles for parallel lines CD and EF)

Thus, $$\triangle DCP \sim \triangle FEP$$ (AAA rule)

Hence, $$\dfrac{DC}{EF} = \dfrac{PC}{EP}$$ (Corresponding sides)

$$\dfrac{27}{EF} = \dfrac{15}{7.5}$$

$$EF = 13.5$$ cm
Question 3
1 / -0

In the following figure, $$XY$$ is parallel to $$BC$$, $$AX= 9 $$ cm, $$XB= 4.5$$ cm and $$BC =18$$ cm. Find $$\displaystyle \dfrac{AY}{YC}$$

A
$$1$$

B
$$4$$

C
$$3$$

D
$$2$$

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)
Hence, $$\dfrac{AB}{AX} = \dfrac{AC}{AY}$$ (Corresponding sides)
On subtracting $$1$$ from both the sides, we have
$$\dfrac{AB}{AX} - 1 = \dfrac{AC}{AY} - 1$$

$$\Rightarrow \dfrac{AB - AX}{AX} = \dfrac{AC - AY}{AY}$$

$$\Rightarrow \dfrac{XB}{AX} = \dfrac{YC}{AY}$$

$$\Rightarrow \dfrac{4.5}{9} = \dfrac{YC}{AY}$$

$$\Rightarrow \dfrac{AY}{YC} = 2$$
Question 4
1 / -0

In the following figure, XY is parallel to $$BC,\ AX= 9 cm$$, $$XB= 4.5$$ cm and $$BC =18$$ cm. Find the value of $$XY$$.

A
$$14$$

B
$$12$$

C
$$16$$

D
$$18$$

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)
Hence, $$\dfrac{AB}{AX} = \dfrac{AC}{AY} = \dfrac{BC}{XY}$$ (Corresponding sides)
$$\dfrac{9 + 4.5}{9} = \dfrac{18}{XY}$$
$$XY = \dfrac{18 \times 9}{13.5}$$
So, $$XY = 12 cm $$
Question 5
1 / -0

For two triangles, if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is called ___ similarity.

A
AAA

B
SSS

C
SAS

D
none of these

Solution

For two triangles, if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is called SAS similarity.
So, option C is correct.
Question 6
1 / -0

In the figure, XY $$\parallel$$ AC and XY divides the triangular region ABC into two equal areas.
Determine AX : XB.

A
$$\displaystyle (\sqrt3-1) : 1$$

B
$$\displaystyle (\sqrt2-1) : 1$$

C
$$\displaystyle (\sqrt7-1) : 1$$

D
$$\displaystyle (\sqrt11-1) : 1$$

Solution

Ar. AXYC $$=$$ Ar. BXY
Ar. AXYC +Ar. BXY $$= 2$$ Ar. BXY
Ar.ABC $$= 2$$ Ar. BXY
Hence, $$\dfrac{Ar. ABC}{Ar. BXY} = 2$$ ...(I)
In $$\triangle$$s ABC and BXY,
$$\angle XBY = \angle ABC $$ (Common angle)
$$\angle BXY = \angle BAC$$ (Corresponding angles of parallel lines XY and AC)
$$\angle BYX = \angle BCA$$ (Corresponding angles of parallel lines XY and AC)
$$\triangle ABC \sim \triangle XBY$$ (AAA postulate)
Now, $$\dfrac{Ar. ABC}{Ar. BXY} = \dfrac{AB^2}{BX^2}$$ (Similar triangle Property)
$$2 = \dfrac{AB^2}{BX^2}$$
$$\dfrac{AB}{BX} = \sqrt{2}$$
$$AB = \sqrt{2} BX$$
$$AX + XB = \sqrt{2} XB$$
$$AX = (\sqrt{2} - 1) XB$$
$$AX : XB = (\sqrt{2} - 1): 1$$
Question 7
1 / -0

In the trapezium ABCD, side AB $$\parallel$$ side DC. Diagonals AC and DB intersect each other at O.
If AB = 15, DC = 10. Find $$\displaystyle \frac{OC}{OA}$$.

A
$$\displaystyle \frac{OC}{OA} = \frac{1}{3}$$

B
$$\displaystyle \frac{OC}{OA} = \frac{7}{3}$$

C
$$\displaystyle \frac{OC}{OA} = \frac{2}{3}$$

D
$$\displaystyle \frac{OC}{OA} = \frac{5}{3}$$

Solution

In $$\triangle OAB$$ and $$\triangle OCD$$,

$$\angle AOB = \angle COD$$ (Vertically opposite angles)

$$\angle OAB = \angle OCD$$ (Alternate angles)

Thus, $$\triangle OAB \sim \triangle OCD$$ (AA rule)

Hence,
$$\dfrac{OC}{OA} = \dfrac{CD}{AB}$$ (C. S. S.T)

$$\dfrac{OC}{OA} = \dfrac{10}{15}$$

$$\dfrac{OC}{OA} = \dfrac{2}{3}$$
Question 8
1 / -0

In parallelogram $$ABCD$$, $$E$$ is mid-point of side $$AB$$ and $$CE$$ meets the diagonal $$BD$$ at point $$O$$, then $$OE : OC $$ is,

A
$$5 \: : \: 2$$

B
$$1 \: : \: 2$$

C
$$9 \: : \: 2$$

D
$$7 \: : \: 2$$

Solution

Given, Parallelogram $$ABCD$$ with $$E$$ as a mid point of $$AB,CE$$ meets diagonal $$BD$$ at $$O$$.

Now, In $$\triangle EOB$$ and $$\triangle DOC$$

$$\angle DOC=\angle EOB\qquad $$ (Vertically opposite angles are equal.)

$$\angle EBO=\angle ODC$$ (Alternate angles, $$\because AB\parallel CD$$ and $$BD$$ intersect it)

$$\angle OCD=\angle OEB$$ (Alternate angles, $$\because AB\parallel CD$$ and $$EC$$ intersect it)

Using $$A{-}A{-}A$$ property, $$\triangle EOB\sim \triangle DOC$$

Using similarity property,

$$\cfrac { OE }{ OC } =\cfrac { EB }{ DC } $$

$$\cfrac { OE }{ OC } =\cfrac { EB }{ AB } $$
   (As $$AB=DC$$,opposite sides of parallelogram)
$$AB=2EB $$
(As $$E$$ is mid point of AB)
$$\therefore \dfrac { OE }{ OC } =\dfrac { 1 }{ 2 } $$

So, $$OE : OC= 1 : 2 $$
Question 9
1 / -0

In the figure, $$\triangle ABC$$ is right angled at $$B$$. $$D$$ is any point on $$AB$$. $$\,seg\,DE\,\bot\, side \,AC$$. If $$AD\,=\,6\,cm,\,AB\,=\,12\,cm\,\,AC\,=\,18\,cm$$. Find $$AE$$.

A
$$5\,cm$$

B
$$3\,cm$$

C
$$4\,cm$$

D
$$6\,cm$$

Solution

In $$\triangle ABC$$ and $$\triangle ADE$$,
$$\angle BAC = \angle DAE$$ (Common angle)
$$\angle ABC = \angle AED$$ (Each $$90^{\circ}$$)
$$\angle ACB = \angle ADE$$ (Third angle)
Thus, $$\triangle ADE \sim \triangle ABC$$ (AAA rule)

Hence, $$\dfrac{AD}{AC} = \dfrac{AE}{AB}$$
$$\Rightarrow \dfrac{6}{18} = \dfrac{AE}{12}$$
$$\Rightarrow AE = \dfrac{6 \times 12}{18}$$
$$\Rightarrow AE = 4$$

Hence, $$AE=4\ cm$$
Question 10
1 / -0

Find BC, if AB = 7.2 cm.

A
$$7.2\ cm$$

B
$$7.1\ cm$$

C
$$0.72\ cm$$

D
$$72\ cm$$

Solution

In $$\triangle ACE$$ and $$\triangle BCD$$

$$\angle ACE = \angle BCD $$ (Common angle)

$$\angle CBD =\angle CAE$$ (Corresponding angles, since $$BD || AE$$)

$$\angle CDB =\angle CEA$$ (Corresponding angles, since $$BD || AE$$)

Hence, $$\triangle ACE \sim \triangle BCD$$ BY AAA similarity criterion.

Thus, $$\dfrac{BC}{AC} = \dfrac{CD}{CE}$$

$$\Rightarrow \dfrac{BC}{BC + AB} = \dfrac{CD}{CD+ DE} $$

Given that, $$CD = DE$$

Hence, $$\dfrac{BC}{BC + AB} = \dfrac{1}{2} $$

$$\Rightarrow 2 BC = AB + BC $$

$$\therefore \ BC = AB $$

$$\Rightarrow BC = 7.2 \ cm $$.

Hence, $$BC=7.2\ cm$$