Question 1: sm6In \triangle ABC, \angle ABC = \angle DAC, AB= 8 \ cm, AC = 4 \ cm \ and \ AD = 5 \ cm .

(i) Prove that \triangle ACD is similar to \triangle BCA .

(ii) Find BC \ and \ CD .

(iii) Find area \ of \  \triangle ACD : area \ of \ \triangle ABC   [2014]

Answer:

(i) In \triangle ACD \ and\  \triangle BCA

\angle C = \angle C (common angle)

\angle ABC = \angle CAD   (given)

\triangle ACD \sim \triangle BCA (AAA postulate)

(ii) Since \triangle ACD \sim \triangle BCA

\frac{AC}{ BC} = \frac{CD}{ CA} = \frac{AD}{ BA}

\frac{4 }{ BC}= \frac{CD}{ 4} = \frac{5}{ 8}

\Rightarrow BC = \frac{8}{ 5} \times 4 = 6.4 \ cm

\Rightarrow CD = \frac{5}{ 8} \times 4 = 2.5 \ cm

(iii) Since \triangle ACD \sim \triangle ABC

Therefore \frac{Area \triangle ACD}{Area \triangle ABC} = \frac{AC^2}{AB^2} = \frac{4^2}{8^2} = \frac{1}{4}

Hence Area \triangle ACD : Area \triangle ABC= 1:4

Answer:

\\

s54Question 2: In the given triangle P, Q \ and \  R are mid points of sides AB, BC \ and \  AC respectively. Prove that \triangle PQR \sim \triangle ABC .

Answer:

In \triangle ABC, PR \parallel BC

Consider \triangle ABC \ and\  \triangle PAR

\angle ABC = \angle APR  (alternate angles)

\angle ARP =\angle ACB  alternate)

Therefore \triangle ABC \sim \triangle ARP (AAA postulate)

\frac{PR}{BC}=\frac{AP}{AB}

Since P is the mid point of AB

AB = 2 AP

\frac{PR}{BC}=\frac{1}{2}

Similarly,  we can prove \frac{PQ}{AC}=\frac{1}{2} and \frac{RQ}{AB}=\frac{1}{2}

Therefore \frac{PR}{BC}=\frac{PQ}{AC}=\frac{RQ}{AB}

Therefore \triangle ABC \sim \triangle PQR  (SSS postulate)

\\

s53.jpgQuestion 3: In the following figure AD \ and \  CE are medians of \triangle ABC. DF \parallel CE . Prove that :

(i) EF=FB

(ii) AG : GD = 2:1

Answer:

(i) Consider \triangle BDF \ and\  \triangle BCE

DF \parallel CE

\angle BDF = \angle BCE  (alternate angles)

\angle BFD = \angle BEC (alternate angles)

Therefore \triangle BDF \sim \triangle BCE (AAA postulate)

\frac{BD}{BC}=\frac{BF}{BE}

\frac{BD}{2BD}=\frac{BF}{BE}

\Rightarrow 2 BF = BE

\Rightarrow BF = FE

(ii)  Consider \triangle AFD \ and \  \triangle AEG

FD \parallel EG

\angle DFA = \angle GEA  (alternate angles)

\angle FDA =\angle EGA (alternate angles)

Therefore \triangle AFD \sim \triangle AEG (AAA postulate)

\frac{AG}{GD}= \frac{AE}{FE} By basic proportionality theorem

Given AE = EB

AE = 2 FE \Rightarrow \frac{AE}{FE} = 2

\frac{AG}{GD}= \frac{2}{1}

\\

Question 4: In the given figure \triangle ABC \sim  \triangle PQR . AM \ and \ PN are altitudes  where as AX \ and \  PY are medians. Prove:

\frac{AM}{PN} = \frac{AX}{PY} 

s52

Answer:

(Since \triangle ABC \sim \triangle PQR

\frac{AB}{PQ}= \frac{AC}{PR} = \frac{BC}{QR}

Given AX \ and \  PY are medians

Therefore 2BX = BC and 2 YR = QR

\frac{AB}{PQ}= \frac{BC}{QR} = \frac{2BX}{2QY} =  \frac{BX}{QY}    … … … … (i)

Consider \triangle ABM \ and\  \triangle PQN

\angle ABN = \angle PQN    Since \triangle ABC \sim \triangle PQR

\angle AMB = \angle PNQ = 90^o (alternate angles)

Therefore \triangle ABM \sim \triangle PQN (AAA postulate)

\frac{AB}{PQ}= \frac{AM}{PN} … … … … (ii)

From (i) and (ii) we get

\frac{AB}{PQ}= \frac{AM}{PN}

and \angle ABX = \angle PQY

Therefore \triangle ABX \sim \triangle PQY

\Rightarrow \frac{AB}{PQ}= \frac{AX}{PY}

Hence \frac{AM}{PN} = \frac{AX}{PY} 

\\

Question 5: Two similar triangles are equal in area. Prove that the triangles are congruent.

Answer:

Let the two triangles be ABC \ and \  PQR

Since the two triangles are similar, We know

\frac{Ar. \ \triangle ABC}{Ar. \ \triangle PQR} = \frac{AB^2}{PQ^2}=\frac{BC^2}{Qr^2}=\frac{AC^2}{PR^2} 

Since the are of the two triangles is equal

\Rightarrow AB=PQ, BC=QR \ and \ AC = PR 

Therefore \triangle ABC \cong \triangle PQR 

\\

Question 6: The ratio between the altitudes of two similar triangles is 3:5 . Write the ratios between their (i) medians (ii) perimeters (iii) areas.

Answer:

The ratio of the altitude of two similar triangles  is the same as the ratio of their sides. Given ratio = 3:5

(i) Ratio between their median = 3:5

(ii) Ratio between their perimeter = 3:5

(iii) Ratio between their areas = 3^2:5^2 = 9:25

\\

Question 7: The ratio between the altitudes of two similar triangles is 16:25 . Find the ration between their: (i) perimeters (ii) altitudes (iii) medians

Answer:

The ratio between the altitudes of two similar triangles is 16:25 .

This means that the ratio of the sides of the triangles = 4:5

(i)  Ratio between their perimeter = 4:5

(ii) Ratio between their altitude = 4:5

(iii) Ratio between their median = 4:5

\\

s51Question 8: The following figure shows a \triangle PQR in which XY \parallel QR . If PX:XQ=1:3 and QR =9 \ cm , find the length of XY .

Answer:

Given PX:XQ=1:3 and QR =9 \ cm

Consider \triangle PXY \ and \  \triangle PQR

\angle PXY = \angle PQR  (alternate angles)

\angle PYX =\angle PRQ (alternate angles)

Therefore \triangle PXY \sim \triangle PQR (AAA postulate)

\frac{PX}{PQ}= \frac{XY}{QR}

\frac{PX}{PX+XQ}= \frac{XY}{QR}

\frac{PX/XQ}{PX/XQ+1}= \frac{XY}{QR}

\frac{1}{4}= \frac{XY}{9}

\Rightarrow XY = \frac{9}{P4} 

\\

Question 9: In the following figure, AB, CD \ and \ EF are parallel lines. AB=6 \ cm, \ CD= y \ cm, \ EF=10 \ cm,  AC = 4 \ cm \ and   \ CF= x \ cm . Calculate: x \ and \ y  [1985]sm5

Answer:

Consider \triangle FDC \ and\  \triangle FBA

\angle FDC = \angle FDA (Corresponding angles)

\angle DFC = \angle BFA (common angle)

\triangle FDC \sim \triangle FBA (AAA Postulate)

Therefore  \frac{CD}{ AB} = \frac{FC}{ FA}

 \frac{y}{ 6} = \frac{x}{ x+4}

Now consider \triangle FCE \ and \  \triangle ACB

\angle FCE = \angle ACB  (Vertically opposite angles)

\angle CFE = \angle CAB (Alternate angles)

\triangle FCE \sim \triangle ACB (AAA postulate)

Therefore  \frac{FC}{ AC} = \frac{EF}{ AB}

x = \frac{10}{ 6} \times 4 = 6.67 \ cm

Also y= \frac{6.67 }{ 6.67+4} \times 6 = 3.75 \ cm

\\

Question 10: On a map, drawn to a scale of 1:20000 , a rectangular plot of land ABCD has  AB = 24 \ cm and BC = 32 \ cm . Calculate:

(i) the diagonal distance  of the plot in km

(ii) the area of the plot in km^2

Answer:

k = \frac{1}{20000}

Length of AB on map = k \times actual length of AB

Actual length of  AB = 24 \times 20000 \ cm = 4.8 \ km

Similarly Actual length of BC = 32 \times 20000 \ cm = 6.4 \ km

(i) Therefore the diagonal = \sqrt{4.8^2+6.4^2} = 8 \ km

(ii) Area of the plot = 4.8 \times 6.4 = 30.72 \ km

\\

Advertisements