Class 10 (All in one) – TrianglesContact Number: 9667591930 / 8527521718

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*PQR* is an equilateral triangle with each side of length 2*p*. If *PS* ⊥ *OR*, then find the value of *PS*.

In the given figure, ∠CAB = 90°, AD ⊥ BC and ∆BDA ~ ∆BAC. If AC = 75 cm, AB = 1 m and BC = 1.25 m, then find the value of AD.

The perimeters of two similar ∆ABC m ∆PQR are respectively 18 cm and 12 cm. if PQ = 5 cm, then find AB.

A vertical stick 1 m long casts a shadow 80 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower.

*ABCD* is a trapezium in which *AB* || *DC*. *P* and *Q* are points on sides *AD* and *BC* respectively such that *PQ* || *AB*. If *PD* = 18 cm, *BQ* = 35 cm and *QC* = 15 cm, find the value of *AD*.

In the given figure, if ∠1 = ∠2 and ∆NSQ = ∆MTR, prove that ∆PTS ~ ∆PRQ.

In the given figure, $\Delta ABC\text{\hspace{0.17em}}=\text{\hspace{0.17em}}{90}^{\circ}$ and $CD\text{\hspace{0.17em}}\perp \text{\hspace{0.17em}}AB$. Prove that $\frac{B{C}^{2}}{A{C}^{2}}=\frac{BD}{AD}.$

The diagonals of a rhombus are 30 cm and 40 cm. Find each side of the rhombus.

In ∆*ABC*, if ∠*ADE* = ∠*B*, then prove that ∆*ADE* ~ ∆*ABC*. Also, if *AD* = 7.6 cm, *AE* = 7.2 cm, *BE* = 4.2 cm and *BC* =8.4 cm, find *DE*.

A girl of height 100 cm is walking away from the base of a lamppost at a speed of 1.9 m/s. If the lamp is 5 m above the ground, find length of her shadow after 4s.

A farmer has a field in the shape of a right angled triangle with perpendicular sides of lengths 128 m and 64 m. He wants to leave a space in the form of a square of largest size inside the field for growing wheat and the remaining for growing vegetables. Find the length of the side of such a squared space. What value is indicated from this action?

In $\Delta PQR,\text{\hspace{0.17em}\hspace{0.17em}}P{R}^{2}-P{Q}^{2}=Q{R}^{2}$ and *M* is a point on side *PR* such that *QM* ⊥ PR. Prove that $Q{M}^{2}=PM\times MR$.

In a ∆*PQR*, *N* is a point on *PR*, such that *QN* ⊥ *PR*. If *PN* · *NR* = *QN*^{2}, prove that ∠*PQR* = 90°.

In right angled D*ABC*, right angled at *C*, *P* and *Q* are points of sides *CA* and *CB* respectively, which divide these sides in the ratio 2:1. Prove that 9*AQ*^{2} = 9*AC*^{2} + 4*BC*^{2}.

*PB* and *QA* are the perpendiculars to segment *AB*. If *PO* = 5 cm, *QO* = 7 cm and ar(∆*BOP*) = 150 cm^{2}, find ar(∆*QOA*).

In a quadrilateral *ABCD*, ∠*A* + ∠*D* = 90°. Prove that *AC*^{2} + *BD*^{2} = *AD*^{2} + *BC*^{2}.

If *S* is a point on side *PQ* of a D*PQR* such that *PS* = *QS* = *RS*, prove that *PR*^{2} + *OR*^{2} = *PQ*^{2}.

In the figure of ∆*ABC*, *P* is the mid-point of *BC* and *‘Q*’ is the middle point of *AP*. If extended *BQ* meets *AC* in *R*, prove that $RA=\frac{1}{3}CA$.

In the given figure, *AP* = 3 cm, *AR* = 4.5 cm, *AQ* = 6 cm, *AB* = 5 cm and *AC* = 10 cm, then find *AD* and ratio of areas of ∆*ARQ* and ∆*ADC*. **CBSE 2015**

In the given figure, *O* is any point inside a rectangle *ABCD* such that *OB* = 6 cm, *OD* = 8 cm and *OA* = 5 cm. Find the length of *OC*.

*ABCD* is a parallelogram in the given figure, *AB* is divided at *P* and *CD* at *Q*, so that *AP* : *PB* = 3 : 2 and *CQ* : *OD* = A : 1. If *PQ* meets *AC* at *R*, prove that $AR=\frac{3}{7}AC$.

In the given figure, ∠*AEF* = ∠*AFE* and *E* is the mid-point of *CA*. Prove that $\frac{BD}{CD}=\frac{BF}{CE}$. **NCERT Exemplar; CBSE 2011**

In the given figure, if *PQRS* is a parallelogram and *AB* || *PS*, prove that *OC* || *SR. ***NCERT Exemplar**

Two trees of heights *x* and *y* are *d m* apart.

(*i*) Prove that the height of the point of intersection of the line joining the top of each tree to the foot of the opposite trees is given by $\frac{xy}{x+y}$ m.

(*ii*) Which mathematical concept is used in this problem?

(*iii*) What are the values depicted here?

Shweta prepared two posters on National Integration for decoration on Independence day on triangular sheets (say *ABC* find *DBF*). The sides *AB* and *AC* and the perimeter *P*_{1} of ∆*ABC* are respectively four times the corresponding sides *DE* and *DF* and the perimeter *P*_{2} of ∆DEF. Are the two triangular sheets similar? If yes, find $\frac{\text{ar\hspace{0.17em}}\left(\Delta ABC\right)}{\text{ar\hspace{0.17em}}\left(\Delta DEF\right)}$. What values ran be indicated through celebration of national festivals?

There is a triangular park *ABC* in a colony (as shown in figure). The Resident Welfare Association of the colony wishes to divide this park into two parts of equal areas—one for planting trees and raising a lawn and the other for providing place for children playing activities. One of the members Salma suggested to draw a line segment *XY* || *BC* for this purpose.

(*i*) State how this line segment *XY* can be drawn, so that *X* and *Y* lie *AB* and *AC* respectively?

(*ii*) What value is depicted from this action?

A man steadily goes 8 m due East and then 6 m due North.

(*i*) Find the distance from initial point to last point.

(*ii*) Which mathematical concept is used in this problem?

(*iii*) What value is indicated in this question?

Consider two similar ∆*ABC* and ∆*PQR* as shown in the following figure.

(*i*) Prove that, if the areas of ∆*ABC* and ∆*PQR* are equal, then the ∆*ABC* and ∆*PQR* are always congruent.

(*ii*) If the sides of two similar triangles are in the ratio 2 : 5, then find the ratio of the areas of these triangles.

(*iii*) If the shape of the two triangles are same but size is different, then both the triangles are congruent. Is this statement true? Justify it.

(*iv*) Suppose, a person wants to select the triangle having maximum area. For maximum area, he has to select the ∆*PQR*. What his decision shows?

Amir was asked to prove the basic proportionality theorem. The proof given by him is as follows:**Given** *ABC* is a triangle and *DE* is drawn parallel to *BC* to intersect *AB* and *AC* at *D* and *E*, respectively.**To prove** $\frac{AD}{DB}=\frac{AE}{EC}$**Proof** *DE* || *BC* [given]

So, ∠1 = ∠2 and ∠3 = ∠4 [corresponding angles]

$\therefore \Delta ADE~\Delta ABC$ [by AAA similarity criterion]

$\Rightarrow \frac{AD}{AB}=\frac{AE}{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{...}\left(i\right)$

$\Rightarrow 1-\frac{AD}{AB}=1-\frac{AE}{AC}$ [multiplying by '–' sign on both sides and then adding 1 on both sides]

$\Rightarrow \frac{AB-AD}{AB}=\frac{AC-AE}{AC}$

$\Rightarrow \frac{DB}{AB}=\frac{EC}{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{...}\left(\text{ii}\right)$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{AD}{AB}\times \frac{AB}{DB}=\frac{AE}{AC}\times \frac{AC}{EC}$

$\therefore \frac{AD}{DB}=\frac{AE}{EC}$

His classmate Madhu told him that his proof is not valid and provided him the correct proof. Amir thanked Madhu for her gentle gesture.

(*i*) Why Amir's proof was not valid?

(*ii*) Write the proof given by Madhu.

(*iii*) What value is depicted from this action?

In the given figure, if ∆ADB ~ ∆ADC, then find the value of *p*.

Find the value of unknown variables, if ∆ABC and ∆PQR are similar.

In ∆PQR and ∆MST, ∠P = 55°, ∠Q = 25°, ∠M = 100° and ∠S = 25°. Is ∆QPR ~ ∆TSM? Why? **NCERT Exemplar**

∆ABC and ∆AMP are two right angled triangles, right angled at B and M, respectively. Prove that CA × MP = PA × BC. **NCERT Exemplar **

If ∆ABC ~ ∆DFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm, then find DE and ∠F. **NCERT Exemplar**

Prove that the line segments joining the mid-points of the sides of a triangle form four triangle, each of which is similar to the original triangle.

In the following figure, if $\frac{AD}{DC}=\frac{BE}{EC}$ and ∠CDE = ∠CED, then prove that ∆CAB is an isosceles triangle.

In the given figure, ∆OAB ~ ∆OCD. If AB= 8 cm, BO = 6.4 cm, OC = 3.5 cm and CD = 5 cm, then find the values of OA and DO.

ABCD is a trapezium with AB || DC. If ∆AED ~ ∆BEC, then prove that AD = BC.

In a ∆ABC, P and Q are points in AB and AC, respectively and PQ || BC. Prove that the median bisects PQ.

In the given figure, if AB || CD, then find the value of x.

In the given figure, D and E are two points lying on side AB, such that AD = BE. If DP || BC and EQ || AC, then prove that PQ || AB. **CBSE 2012,10**

In the given figure, PA, QB, RC and SD are all perpendiculars to a line *l*. AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS. **NCERT Exemplar**

State whether the given pairs of triangles are similar or not. In case of similarity, mention the criterion. **CBSE 2015**

In the given figure, if ∠BAC = 90° and AD ⊥ BC, prove that AD^{2} = BD · CD. **NCERT Exemplar**

A 15 m high tower casts a shadow 24 m long at a certain time and at the same time, telephone pole casts a shadow 16 m long. Find the height of the telephone pole. **NCERT Exemplar**

In the given figure, ∠M = ∠N = 46°. Express x in terms of a, b and c, where a, b and c are the lengths of LM, MN and NK, respectively. **CBSE 2009**

P and Q are the points on sides AB and AC, respectively of ∆ABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3PQ. **CBSE 2010**

A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3 m, find how far is she away from the base of the pole? **NCERT Exemplar**

In the adjoining figure, ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4cm and CD = 5 cm, find BD and AB. **NCERT Exemplar**

In the given figure, if AB || DC and AC, PQ intersect each other at the point O, then prove that OA · CQ = OC · AP. **NCERT Exemplar**

In the given figure, *l *|| *m* and line segments AB, CD and EF are concurrent at point P. Prove that

$\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$ **NCERT Exemplar**

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of any equilateral triangle described on one of its diagonals.

Equilateral triangles are drawn on the sides a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

AD is an altitude of an equilateral ∆ABC. In AD as base another equilateral triangle ADE is constructed. Prove that ar (∆ADE); ar(ABC) = 3 : 4. **CBSE 2010**

A ladder 17 m long, reaches at a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

If the sides of a triangle are 3 cm, 4 cm and 6 cm long, then determine whether the triangle is a right angled triangle.

In the given figure, ∠B < 90° and segment AD ⊥ BC. Show that

(i) b^{2} = h^{2 }+ a^{2 }+ x^{2} – 2ax

(ii) b^{2} = a^{2} + c^{2} – 2ax

If the lengths of the diagonals of rhombus are 16 cm and 12 cm. Then, find the length of the sides of the rhombus. **NCERT Exemplar**

In the given figure, if CD = 17 m, BD = 8 m and AD = 4 m, find the value of AC. **CBSE 2014**

Find the altitude of an equilateral triangle of side 8 cm. **NCERT Exemplar**

In ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, then prove that (a + b) (a – b) = (c + d) (c – d). **NCERT Exemplar**

∆ABC is a right triangle in which ∠C = 90° and CD ⊥ AB. lf BC = a, CA = b, AB = C and CD = p, then prove that

(i) cp = ab.

(ii) $\frac{1}{{p}^{2}}=\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}$ **CBSE 2011**

D is a point on the side BC of an equilateral triangle. ABC, such that DC = $\frac{1}{4}$BC. Prove that AD^{2} = 13CD^{2}.

In ∆ABC, ∠C is an obtuse angle, AD ⊥ BC and AB^{2} = AC^{2} + 3 BC^{2}. Prove that BC = CD.

There is a staircase as shown in figure connecting points A and B. Measurements of steps are marked in the figure. Find the straight distance between A and B.

For going to city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway, which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction on the highway? **NCERT Exemplar**

Find the third side of a right angled triangle whose hypotenuse is of length p cm, one side of length q cm and p – q = 1. **CBSE 2015**

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