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11000 , 11001 , 11002

11000 , 11001 , 11002 are the three natural numbers after 10999

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR.Show that Δ ABC ~ Δ PQR

Answer:To Prove:Δ ABC∼Δ PQRGiven:Proof:

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to

E, Q to L, and R to L

We know that medians divide opposite sides.

Hence, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC,Diagonals AE and BC bisect each other at point D.

Therefore,Quadrilateral ABEC is a parallelogram.

AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given in the question that,







ΔABEΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal

∠BAE =∠QPL (i)

Similarly, it can be proved that

ΔAECΔPLR and

∠CAE =∠RPL (ii)

Adding equation (i) and (ii), we obtain

∠BAE +∠CAE =∠QPL +∠RPL

⇒∠CAB =∠RPQ (iii)

In ΔABC and ΔPQR,

(Given)

∠CAB =∠RPQ [Using equation (iii)]

ΔABCΔPQR (By SAS similarity criterion)

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Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that AB/PQ = AC/PR = AD/PM

To Prove: ΔABC ~ ΔPQR

Construction: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

Proof: In ΔABD and ΔCDE, we have

AD = DE [By Construction]

BD = DC [∴ AP is the median]

and, ∠ADB = ∠CDE [Vertically opp. angles]

∴ ΔABD≅ΔCDE [By SAS criterion of congruence]

⇒ AB = CE [CPCT] ...(i)

Also, in ΔPQM and ΔMNR, we have

PM = MN [By Construction]

QM = MR [∴ PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [By SAS criterion of congruence]

⇒ PQ = RN [CPCT] ...(ii)

Now, AB/PQ = AC/PR = AD/PM

⇒ CE/RN = AC/PR = AD/PM ...[From(i)and(ii)]

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [∴ 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [By SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1+∠2=∠3+∠4

⇒ ∠A = ∠P ...(iii)

Now, In ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Given)

∠A = ∠P [From(iii)]

∴ ΔABC ~ ΔPQR [By SAS similarity criterion]

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rational number should be 333

5 + 3 , 4 +4, 3+5 are probabilitys of getting a sum of 8

It is given that ΔABC ~ ΔPQRWe know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR ...(i)Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 ...(iii)From equations(i)and(iii), we getAB/PQ = BD/QM ...(iv)In ΔABD and ΔPQM,∠B = ∠Q [Using equation(ii)]AB/PQ = BD/QM [Using equation(iv)]∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM

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can you please post a clearer image of the question?

Extended area=66*46-60*40=636

11,00011,00111,002

AREA OF PLAYGROUND = 80*60=4800Now, According to question cemented pathway outside ittherefore, LENGTH WILL BE= 80+2+2=84 BREADTH WILL BE =60+2+2=64Now AREA OF PATH INCLUDING PLAYGROUND=84*64=5376hence area of cemented path only = 5376-4800=576cost of cementing =576*20=11520

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM ofΔ PQR (see Fig. 6.41). Show thatΔ ABC ~ Δ PQR.

Answer:To Prove:Δ ABC∼Δ PQRGiven:

Proof:Median divides the opposite sideBD =and,

QM =

Now,Multiplying and dividing by 2, we get,=



In Δ ABD and Δ PQM,

Side-Side-Side (SSS) Similarity Theorem - If the lengths of thecorrespondingsides of two triangles areproportional, then the triangles must besimilar.ΔABDΔPQM (By SSS similarity)

∠ABD =∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD =∠PQM (Proved above)

The SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both arecongruent, then the two triangles are similar.ΔABCΔPQR (By SAS similarity)Hence, Proved.

If all the three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent by SSS rule.

In above given figure, AB= PQ, QR= BC and AC=PR, hence ∆ ABC ≅ ∆ PQR.

Answer: B) Rs. 13, Rs. 17

Explanation:

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13.

We cannot ocnstruct a triangle if we are given only three angles.

two angles and one side

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are you in 8 class tell mee

The formula of perimeter of a kite is 2×(l+b)Here we have its length given that is 23cm, so as per the question-(we need to convert m into cm so 1m=100cm, 106×100)2×(23+b)=10600cm56+2b=10600cm2b=10600-562b=10544cmb=10544/2b=5272cmAnd now as we all know there are two pairs of equal sides in a kite so there is a pair of 23cm and 5272cm ANS.

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in throw of 2 dice

total outcomes=36

i) outcomes of duplets= (1,1)(2,2)(3,3)(4,4)(5,5)(6,6)=6

probability=6/36=1/6

ii)outcomes of sum of 11= (6,5)(5,6)=2

probability =2/36=1/18

Route x +Y=11 x+route y=7

When we throw two die we have 6² i.e 36possiblity. So n(S) = 36i)Let the event A be having prime on each diceSo chances are {(2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)}Hence n(A) = 9Thus P(A) = 9/36= 1/4ii)

Let B be the event having the sum of 9 or 11So chances will be like {(3,6), (6.3), (4.5), (5,4), (5,6), (6,5)}So n(B) = 6Thus P(B) = 6/36 =1/6

length of picture is = 16.1 cmand breadth of picture is = 10.4 cm.

length of pictures is =16.1cm and breadth of picture is 10.4 cm.

Length=18.5-1.2=17.3cmBreadth=12.8-1.2=11.6Perimeter of picture=2*(17.3+11.6)2*28.9=57.8cm

11.2km is correct answer

Thesix elements of a triangleare its three angles and the three sides. The line segment joining a vertex of atriangleto the mid point of its opposite side is called a median of thetriangle. Atrianglehas 3 medians.

Thesix elements of a triangleare its three angles and the three sides. The line segment joining a vertex of atriangleto the mid point of its opposite side is called a median of thetriangle. Atrianglehas 3 medians.

Sample space : 1, 2, 3, 4, 5, 6

Total possibilities = 6

i) A=getting 8 = zero possiblity

P(A) = 0/6 = 0

ii) B = a number greater than 8 = zero possiblity

P(B) = 0/6 = 0

iii) C = number less than 8 = {1, 2, 3, 4, 5, 6} P(C) = 6/6 = 1

4ab + 6ac + 2bx + 3cx = 2a(2b+3c) + x(2b+3c) = (2a+x)(2b+3c)

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If 2a²b*4ab² we get the value=8a³b³

answer of this question 8a3b3

2a^2b × 4a^2b =8a^4b^2this is right answer

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

SOLUTION:

In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1)

[Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ

QS/QR = PQ/QT…... …(ii) [Taking reciprocals]

[From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Hence, proved

Suppose 8A = 12B = 6C = k A = k/8 B = k/12 C = k/6 A : B : C = k/8 : k/12 : k/6 = 6 : 4 : 8 = 3 : 2 : 4

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i-eii-diii-biv-cv-a

Take LCMcos²A+(1+sinA)²/(1+sinA)cosA=cos²A+1+sin²A+2sinA/(1+sinA)cosA2+2sinA/(1+sinA)cosA2(1+sinA)/(1+sinA)cosA=2/cosA=2SecA

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X2 +2x +2 is a factor of

=x(x+2)+1(x+2)=x=-2or-1f(-2)=16-8-12-8+2=-10so it is nota factor

When quantities are in continued proportion,allthe ratios areequal. If a:b = b:c = c:d = d:e,

As m= tan theeta

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