# Extra questions for Similar triangles

Given below are the Similar Triangles Class 10 Extra Questions.
a. Calculation problems
b. True & False Questions
c. Multiple Choice Questions
d. Proof Questions

Question 1.
P and Q are the mid-point of the sides CA and CB respectively of a $\Delta ABC$, right angled at C. Prove that:
(i) $4AQ^2 = 4AC^2 + BC^2$
(ii)$4BP^2 = 4BC^2 + AC^2$
(iii) $4(AQ^2 + BP^2) = 5AB^2$.

In$\Delta ACQ$
By pythagorus theorem
$AQ^2 = AC^2 + QC^2$
Now $QC = \frac {BC}{2}$

$AQ^2 = AC^2 + (\frac {BC}{2})^2$

$AQ^2 = AC^2 + \frac {BC^2}{4}$
or
$4AQ^2 = 4AC^2 + BC^2$ --(1)
Hence (i) proved

Now In$\Delta BPC$
$BP^2 = BC^2 + PC^2$
Now
$PC= \frac {AC}{2}$
$BP^2 = BC^2 + (\frac {AC}{2})^2$
$BP^2 = BC^2 + \frac {AC^2}{4}$
$4BP^2 = 4BC^2 + AC^2$ --(2)
Hence (ii) proved

$4(AQ^2 + BP^2)= 4AC� +BC^2 + AC^2 +4BC^2$
$4(AQ^2 + BP^2) = 5(AC^2 + BC^2)$
Now in $\Delta ABC$
$AC^2 + BC^2 = AB^2$

Hence
$4(AQ^2 + BP^2) = 5(AB^2)$

Question 2.
Prove that the areas of two similar triangles are in the ratio of the squares of the corresponding altitude
Question 3.
"The areas of two similar triangles are in the ratio of the squares of the corresponding median.". Prove it
Question 4.
"The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments".Prove it
Question 5.
If, AD BE and CF are medians of the $\Delta ABC$, then prove that
$3 (AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$

We will need to some construction to apply pythagorus theorem and get the prove then
a. Draw AM perpendicular to side BC
b. Draw BN perpendicular to side AC
c. Draw CK perpendicular to side AB

Now it is given that
$BD=DC= \frac {BC}{2}$
$FB=FA= \frac {AB}{2}$
$AE=EC= \frac {AC}{2}$

Now in $\Delta ABM$ and $\Delta AMC$
$AB^2 = AM^2 + BM^2$
$AC^2 = AM^2 + CM^2$
Now BM= BD- MD= DC-MD
and CM= DC + MD
Substituting those values and add equations
$AB^2+ AC^2= 2AM^2 + (DC-MD)^2 + (DC + MD)^2$
$AB^2+ AC^2= 2AM^2 + DC^2 + MD^2 -2 DC .MD + DC^2 + MD^2 +2 DC .MD$
$AB^2+ AC^2= 2AM^2+ 2DC^2 + 2 MD^2$
$AB^2+ AC^2=2( AM^2 + MD^2 ) + 2 (\frac {BC^2}{4}$
Now from pythagorus theorem in triangle AMD
$AM^2 + MD^2= AD^2$
So,
$AB^2+ AC^2=2 AD^2 + \frac {BC^2}{2}$
Similary we can prove the same equation for the triangles $\Delta BKC$ and $\Delta AKC$
$AB^2+ BC^2=2 AE^2 + \frac {AC^2}{2}$
Similarly for the triangle $\Delta ABN$ and $\Delta BNC$

$AC^2+ BC^2=2 AF^2 + \frac {AB^2}{2}$
$3 (AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$

Question 6.
In an equilateral triangle ABC, D is a point on side BC such that $BD = \frac {1}{3} BC$. Prove that $9AD^2 = 7AB^2$

Let draw a perpendicular to line AM to BC

Now we know that Perpendicular bisect the line in equilateral triangle.So
BM= MC

Now in $\Delta ABM$
$AB^2 = BM^2 + AM^2$
$AB^2 = \frac {BC^2}{4} + AM^2$ -(1)
Now in $\Delta AMD$
$AD^2 = AM^2 + MD^2$
Now $MD = BM - BD = \frac {BC}{2} - \frac {BC}{3}= \frac {BC}{6}$
So
$AD^2 =AM^2 + \frac {BC^2}{36}$ --(2)
Subtracting (1) and (2)
$AB^2 - AD^2 = \frac {BC^2}{4} - \frac {BC^2}{36}$
$AB^2 - AD^2=\frac {2BC^2}{9}$
$9AB^2 - 9 AD^2 = 2BC^2$
Now BC=AB
$9AD^2 = 7AB^2$

Question 7.
O is a point in the interior of a triangle ABC, $OD \perp BC$, $OE \perp AC$ and $OF \perp AB$. Show that:
a. $OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2= AF^2 + BD^2 + CE^2$.
b. $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$

The Figure is shown below

Join the points O and A, O and B, O and C

Now in $\Delta OAE$
$OA^2 = OE^2 + AE^2$ -(1)

Now in $\Delta ODC$
$OC^2 = CD^2 + OD^2$ -(2)
Now in $\Delta OBF$
$OB^2 =OF^2 + BF^2$ -(3)
$OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2=AE^2 + CD^2 + BF^2$ -(A)

Now in $\Delta OEC$
$OC^2 = OE^2 +CE^2$ -(4)
Now in $\Delta OBD$
$OB^2 = OD^2 + BD^2$ --(5)
Now in $\Delta OAF$
$OA^2 = Af*2 + OF^2$ --(6)
$OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2=AF^2 + BD^2 + CE^2$ -(B)
Hence (a) is proved
From equation (A) and (B)
$AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$
Hence (b) is proved

Question 8.
D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB.CD$

In $\Delta ADC$ and $\Delta BAC$
$\angle ADC = \angle BAC$
$\angle ACD = \angle BCA$ (Common angle)
Therefore By AA similarity criterion
$\Delta ADC \sim \Delta BAC$
We know that corresponding sides of similar triangles are in proportion.
So
$\frac {CA}{CB} ={CD}{CA}$
$CA^2= CB.CD$

Question 9.
D, E and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$. Find the ratio of the areas of $\Delta DEF$ and $\Delta ABC$

Diagram is shown as below

In $\Delta ADF$ and $\Delta ABC$
$\frac {AD}{AB} = \frac {AF}{FC} = \frac {1}{2}$ ( As D and F are midpoints)
By converse of Basic proportionality theorem
$\Delta ADF \sim \Delta ABC$
and $\frac {DF}{BC } = \frac {1}{2}$ -(1)

In $\Delta DBE$ and $\Delta ABC$
$\frac {DB}{AB} = \frac {BE}{BC} = \frac {1}{2}$ ( As D and E are midpoints)
By converse of Basic proportionality theorem
$\Delta DBE \sim \Delta ABC$
and $\frac {DE}{AC} = \frac {1}{2}$ -(2)

In $\Delta FCE$ and $\Delta ACB$
$\frac {FC}{AC} = \frac {CE}{CB} = \frac {1}{2}$ ( As E and F are midpoints)
By converse of Basic proportionality theorem
$\Delta FCE$ \sim \Delta ACB$and$ \frac {EF}{AB} = \frac {1}{2}$-(3) From (1) ,(2) and (3)$\frac {DF}{BC }=\frac {DE}{AC}=\frac {EF}{AB}=\frac {1}{2}$By SSS criterio$\Delta DEF \sim \Delta CAB $Now Area of the similar triangle is given by$ \frac {area \Delta DEF}{ area \Delta ABC} = \frac {DF^2}{BC^2} = \frac {1}{4}$Question 10. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If$AD \perp BC$and$EF \perp AC$, prove that$\Delta ABD \sim \Delta ECF$Answer It is given that ABC is an isosceles triangle. Hence AB = AC$ \angle ABC = \angle ACB \angle ABD = \angle FCE$-(1) Now In$\Delta ABD$and$ \Delta ECF\angle ADB = \angle EFC$(Each 90°) From (1)$ \angle ABD = \angle FCE$By using AA similarity criterion Therefore$\Delta ABD \sim \Delta ECF$Question 11. O is any point inside a rectangle ABCD. Prove that$OB^2 + OD^2= OA^2 + OC^2$. Answer Draw a line MN passing through O and parallel to AB and CD Now ABCD is a rectangle, then ABNM and MNCD are all rectangles Now in$\Delta MODOD^2 = OM^2 + MD^2$In$ \Delta OBNOB^2 = BN^2 + ON^2$Adding these$OB^2 + OD^2=OM^2 + MD^2 + BN^2 + ON^2$Now BN= AM and MD=NC$OB^2 + OD^2=(OM^2 + AM^2) + (ON^2 + NC^2)$-(1) In triangle AOM$OA^2 = AM^2 + OM^2$-(2) In Triangle ONC$OC^2 = ON^2 + NC^2$-(3) From (1) ,(2) and (3)$OB^2 + OD^2= OA^2 + OC^2$Question 12. ABCD is a rectangle. Points M and N are on BD such that$AM \perp BD$and$CN \perp BD$. Prove that$BM^2 + BN^2 = DM^2 + DN^2$Answer In$\Delta ADMAD^2 = DM^2 + AM^2$-(1) In$\Delta DNCCD^2 = DN^2 + CN^2$-(2) Adding (1) and (2)$AD^2 + CD^2=DM^2 + AM^2 + DN^2 + CN^2$-(A) In$\Delta AMBAB^2 = AM^2 + BM^2$-(3) In$\Delta CNB BC^2 = BN^2 + CN^2$-(4) Adding (3) and (4)$AB^2 + BC^2 =AM^2 + BM^2 + BN^2 + CN^2$-(B) Now AD= BC and AB = CD$ ,putting in equation (B)
$AD^2 + CD^2=AM^2 + BM^2 + BN^2 + CN^2$ --(C)
Subtracting (A) and (C)
$BM^2 + BN^2 = DM^2 + DN^2$

Question 13.
True and False
(a)All Similar triangles are congureent triangles
(b) All the equilaterals triangles are similar
(c)If the triangle ABC, AB = 5 cm, BC = 12 cm and AC = 13 cm. Then triangle a right triangle
(d) If the Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle,Then the two triangles are similar
(e) If the Sides of two similar triangles are in the ratio 5 : 11. Areas of these triangles are in the ratio 121 : 25

(a) False
(b) True
(c) Here $13^2=12^2 + 5^2$ hence True
(d) True , as the corresponding two sides and the perimeters are three, their third sides will also be three times
(e) False as the Areas of these triangles are in the ratio 25: 121

Question 14.
Two triangles are similar, and the ratio of their corresponding sides is 3:5. If the length of one of the sides of the smaller triangle is 9 cm, find the length of the corresponding side of the larger triangle.
(a) 15 cm
(b) 5 cm
(c) 8 cm
(d) 11 cm

Question 15.
In a triangle ABC, the bisectors of angles A and B meet at O. If angle AOB is 120 degrees, then the angle C.
(a) 30 degrees
(b) 60 degrees
(c) 45 degrees
(d) None of these

Question 16.
Which of these is not Pythagoras triplet?
(a) 3,4,5
(b) 5,24,25
(c) 6 ,2.5 ,6.5
(d) None of these

Question 17.
The ratio of their perimeters of two similar triangles is 5:8. If the length of one of the sides of the smaller triangle is 4 cm, find the length of the corresponding side of the larger triangle.
(a) 5 cm
(b) 6.4 cm
(c) 7 cm
(d) 6.5 cm

Question 18.
In triangles ABC and DEF, $\angle B = \angle E$, $\angle F = \angle C$ and AB = 3 DE. Then, the two triangles are
(a) congruent but not similar
(b) congruent as well as similar
(c) neither congruent nor similar
(d) similar but not congruent

(14) Let the corresponding side of the larger triangle be x cm. Then, we have 9/x = 3/5, which gives x = 15 cm. So, option (b) is correct
(15) Let angles A and B be x and y degrees, respectively. Then, we have in AOB
$\frac {x}{2} + \frac {y}{2} + 120 =180$
or x+ y=120
Now in Triangle ABC
x + y + C = 180
Therefore C=60 degrees
Hence option(b) is correct
(16) (b)
(17) The ratio of their corresponding sides is the same as the ratio of their perimeters, which is 5:8. Hence, the length of the corresponding side of the larger triangle is (8/5)*4 = 6.4 cm.
Hence (b) is correct
(18) (d) is correct answer as they are similar but not congruent as sides are not equal

## Summary

This Class 10 Maths Extra questions for Similar triangles with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

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### Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20