Question 11: The dimension of a model of a multi storied building are 1  \ m by 60 \ by 1.20 \ m .  If the scale factor is 1:50 , find the actual dimensions of the building. Also find:

(i) the floor area of a room of the building, if the floor are of the corresponding room in the model is 50 \ cm^2

(ii) the space inside the room of them model if the space inside the corresponding room if the building is 90 \ m^3

Answer:

Dimension of model = 100 cm \times 60 \ cm \times 120 \ cm

Scale factor (k) = \frac{1}{50}

Actual length = 100 \times 50 = 50 m

Actual breadth = 60 \times 50 = 30 m

Actual height = 120 \times 50 = 60 m

Therefore the actual dimension of the building = 50 \ m \times 30 \ m \times 60 \ m

(i) Floor area = 50^2 \times 50 cm^2 = 125000 \ cm^2 = 12.5 \ m^2

(ii) Volume of the model = \frac{90 \ m^3}{50^3} = 720 \ cm^3

\\

sm4Question 12: In \triangle PQR, L and M are two points on the base QR , such that \angle LPQ = \angle QRP and \angle RPM = \angle RQP . Prove that:

(i) \triangle PQL \sim \triangle RPM

(ii) QL \times RM = PL \times PM

(iii) PQ^2= QR \times QL .   [2003]

Answer:

(i)   Consider \triangle PQL \ and \  \triangle RMP

\angle LPQ = \angle QRP (Given)

\angle RQP = \angle RPM (Given)

\triangle PQL \sim \triangle RMP (AAA postulate)

(ii) Since \triangle PQL \sim \triangle RMP 

 \frac{PQ}{ RP} =  \frac{QL}{ PM }=  \frac{PL}{ RM}

\Rightarrow QL \times RM = PL \times PM

(iii) Consider  \triangle PQL \ and \  \triangle RQP

\angle LPQ = \angle QRP (Given)

\angle Q (common angle)

\triangle PQL \sim \triangle RQP (AAA postulate)

Therefore  \frac{PQ}{ RQ} =  \frac{QL}{ QP} =  \frac{PL}{ PR}

\Rightarrow PQ^2 = QR \times QL 

\\

Question 13: In \triangle ABC, \angle ACB = 90^o and CD \perp AB . Prove that: \frac{BC^2}{AC^2}=\frac{BD}{AD} .s62.jpg

Answer:

Consider \triangle ACD \ and \  \triangle ABC

\angle DAC = \angle BAC (Common)

\angle CDA = \angle ACB = 90^o (Given)

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

\frac{AC}{AB}=\frac{AD}{AC}

AC^2=AD \times AB   … … … … (ii)

Consider \triangle BCD \ and \  \triangle ABC

\angle CBD = \angle CBA (Common)

\angle BDC = \angle ACB = 90^o (Given)

\triangle BCD \sim \triangle ABC (AAA postulate)

\frac{BC}{AB}=\frac{BD}{BC}

BC^2=BD \times AB … … … … (ii)

From (i) and (ii)

\frac{BC^2}{AC^2}=\frac{BD \times AB}{AD \times AB} 

Hence Proved.

\\

Question 14: A \triangle ABC with AB=3 \ cm, BC = 6 \ cm \ and \ AC = 4 \ cm is enlarged to a \triangle DEF such that the longest side of \triangle DEF = 9 \ cm . Find the scale factor and hence, the lengths of the other sides of \triangle DEF .

Answer:

Scale factor (k) = \frac{EF}{BC}=\frac{9}{6}= 1.5 

Therefore \frac{ED}{AB}= 1.5 \Rightarrow ED = 3 \times 1.5 = 4.5 \ cm 

\frac{DF}{AC}= 1.5 \Rightarrow ED = 4 \times 1.5 = 6 \ cm 

\\

Question 15: Two isosceles triangles have equal vertical angles. Show that the triangles are similar.  If the ratios between the areas of these two triangles is 16:25 , find the ratio between their corresponding altitudes.

Answer:

Consider \triangle ABC \ and \  \triangle DEF

\angle BAC = \angle EDF (Common)

AB=AC \ and \ DE = DF (Given)

\triangle ABC \sim \triangle EDF (SAS postulate)

\frac{Ar. \ \triangle ABC}{Ar. \ \triangle DEF}=\frac{AD^2}{PS^2}

\frac{16}{25} = (\frac{AD}{PS})^2 

\Rightarrow \frac{AD}{PS}=\frac{4}{5} 

\\

Question 16: In \triangle ABC, AP:PB=2:3. PQ \parallel BC and is extended to Q so that CQ \parallel BA .  Find:

(i) area \ of \ \triangle APO : area \ of \  \triangle ABC

(ii) area \ of  \ \triangle APO : area \ of \  \triangle CQO

Answer:

(i) Consider \triangle APO \ and \  \triangle ABC

\angle APO = \angle ABC (alternate angles)

\angle AOP = \angle ACB  (alternate angles)

\triangle APO \sim \triangle ABC (AAA postulate)

\frac{AP}{PB}=\frac{2}{3}

or \frac{AP}{AB}=\frac{2}{5}

\frac{area \ of \ \triangle APO}{area \ of \  \triangle ABC} =\frac{2^2}{5^2}=\frac{4}{25}

(ii) Consider \triangle APO \ and \  \triangle QOC

\angle AOP = \angle QOC (vertically opposite angles)

\angle PAO = \angle OCQ (alternate angles)

\triangle APO \sim \triangle QOC (AAA postulate)

\frac{area \ of \ \triangle APO}{area \ of \  \triangle QOC} =\frac{AP^2}{CQ^2}= \frac{AP^2}{PB^2}=\frac{2^2}{3^2}=\frac{4}{9}

\\

s61.jpgQuestion 17:The following figure shows a \triangle ABC in which AD \perp BC and BE \perp AC . Show that:

(i) \triangle ADC \sim \triangle BEC

(ii) CA \times CE = CB \times CD

(iii) \triangle ABC \sim \triangle DEC

(iv) CD \times AB = CA \times DE

Answer:

(i) Consider \triangle ADC \ and \  \triangle BEC

\angle ADC = \angle BEC = 90^o (alternate angles)

\angle ADC = \angle BCE  (common angle)

\triangle ADC \sim \triangle BEC (AAA postulate)

(ii) Therefore \frac{CA}{CB}=\frac{CD}{CE}

CA \times CE = CB \times CD

(iii) Consider \triangle ABC \ and \  \triangle DEC

\angle ADC = \angle BCE  (common angle)

\frac{CA}{CB}=\frac{CD}{CE}

\frac{CA}{CD}=\frac{CB}{CE}

\triangle ABC \sim \triangle DEC (SAS postulate)

(iv) Therefore \frac{AC}{DC}=\frac{AB}{DE}

AC \times DE = CD \times AB

\\

Question 18: sm3In the given figure, ABC is a triangle with \angle EDB = \angle ACB . Prove that \triangle ABC \sim \triangle EBD . If BE=6 \ cm, EC = 4 \ cm, BD = 5 \ cm and area of \triangle BED = 9 cm^2 . Calculate the:

(i) length of AB

(ii) area of \triangle ABC    [2010]

Answer:

Consider \triangle ABC \ and \  \triangle EBD

\angle EDB = \angle ACB (given)

\angle DBE = \angle ABC (common)

Therefore \angle DEB = \angle BAC

\triangle ABC \sim \triangle EBD    (AAA postulate)

(i) Given BE=6\ cm, EC=4\ cm, BD=5\ cm

\frac{AB}{ EB} = \frac{BC}{ BD} = \frac{AC }{ED}

AB = \frac{BE+EC }{5} \times 6 = 2 \ cm

(ii)  \frac{Area \ of \ \triangle ABC}{Area \ of \ \triangle EBD}=\frac{AB^2}{EB^2}=\frac{144}{36}

Area of  \triangle ABC = \frac{144}{ 36} \times 9 = 36 \ cm^2

\\

Question 19: sm2In the given figure \triangle ABC is a right angled triangle with \angle BAC = 90^o .

(i) Prove \triangle ADB \sim \triangle CDA

(ii) If BD = 18 \ cm \ and \ CD = 8 \ cm , find AD

(iii) Find the ratio of the area of \triangle ADB is to area of \triangle CDA  [2011]

Answer:

(i)   Let \angle DAB = \theta

Therefore \angle DAC = 90^o - \theta

\angle DBA = 90^o - \theta

\angle DCA = \theta

Therefore \triangle ADB \sim \triangle CDA (AAA postulate)

(ii) \frac{CD}{AD}=\frac{AD}{BD}

\Rightarrow AD^2 = CD \times BD = 8 \times 18 = 144

Therefore AD = \sqrt{144} = 12

(iii) \frac{Area \ of \ \triangle ADB}{ Area \ of \ \triangle CDA} = \frac{\frac{1}{2} AD \times BD} {\frac{1}{2} AD \times CD} = \frac{BD}{CD} = \frac{18}{8}= \frac{9}{4}

\\

Question 20: sm1In the given figure AB  and DE  are perpendiculars to BC .

(i) Prove that \triangle ABC \sim \triangle DEC 

(ii) If AB =  6 \ cm, DE = 4 \ cm  and AC = 15 \ cm , calculate CD 

(iii) Find the ratio of the area \ of  \ \triangle ABC : area \ of \ \triangle DEC  [2013]

Answer:

(i) From  \triangle ABC \ and \   \triangle DEC

\angle ABC = \angle DEC = 90^o (given)

\angle ACB = \angle DCE (common angle)

 \triangle ABC \sim  \triangle DEC (AAA postulate)

(ii) Since  \triangle ABC \sim \triangle DEC

In \triangle ABC \ and \   \triangle DEC ,

\frac{AB}{ DE} = \frac{AC}{ CD}

AB =  6 \ cm, DE = 4 \ cm  and AC = 15 \ cm

Therefore CD = \frac{15}{6} \times 4 = 10  \ cm

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

\frac{Area \ of \   \triangle ABC}{Area \ of \  \triangle DEC} = \frac{AB^2}{ DE^2} = \frac{6^2}{4^2}= \frac{9}{4}

Therefore the Area \ of \  \triangle ABC : Area \ of \  \triangle DEC = 9:4

\\

Advertisements