Extra questions for Similar triangles|Class 10 Maths
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) $4(AQ^2 + BP^2) = 5AB^2$. Solution
In $\Delta ACQ
$
By pythagorus theorem
$AQ^2 = AC^2 + QC^2$
Now $QC = \frac {BC}{2}$
Question 2. 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. Question 4. The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments. Question 5. If, AD BE and CF are medians of the $ \Delta ABC$, then prove that
$3 (AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$ Solution
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 $\Delta BNC$
Question 6.
In an equilateral triangle ABC, D is a point on side BC such that $BD = \frac {1}{3} BC$. Prove that $9AD^2 = 7AB^2$ Solution
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. $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$ Solution
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)
Adding 1,2,3
$OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2=AE^2 + CD^2 + BF^2$ -(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)
Adding 4,5,6
$OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2=AF^2 + BD^2 + CE^2$ -(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 $CA^2 = CB.CD$ Solution
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$ Solution
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 $\Delta ABD \sim \Delta ECF$ Solution
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 $OB^2 + OD^2= OA^2 + OC^2$. Solution
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 $\Delta MOD$
$OD^2 = OM^2 + MD^2$
In $ \Delta OBN$
$OB^2 = BN^2 + ON^2$
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 $BM^2 + BN^2 = DM^2 + DN^2$ Solution
