b) What is the area of the shaded region? Geometry 1 Assignment - Solutions

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1 Geometry 1 Assignment - Solutions 1. ABCD is a square formed by joining the mid points of the square PQRS. LKXS and IZXY are the squares formed inside the right angled triangles DSC and KXC respectively. The side of the square PQRS is b cm. b) What is the area of the shaded region? Area of shaded region = Area of PQRS (Area of ABCD + Area of LKXS + Area of IZYX + Area of AQB) = b 2 - [ b2 2 + b b b2 8 ] = b 2 - [ b2 ] = b2 a) What is the ratio of areas of PQRS and LKXS? a) 4:1 b) 8:1 c) 16:1 d) 12:1 The side of square LKXS = 1 side of 4 square PQRS. Let side of LKXS = a= b In a square ABCD, point M lies on side AB and point N lies on side CD such that MB = 2cm and ND = 1 cm, Line RS is drawn parallel to side BC and PQ is drawn parallel to side AB. Point P is on side AD, point Q is on side BC, point R is on side AB, and point S is on side CD of the square and point O is the intersection point of the lines PQ, RS and MN. If ONS = 40 0, and MSC = 60 0, what is the value of OMS? a) 20 b) 30 c) 40 d) None of these Then area of LKXS = a 2 = ( b 4 )2 = b2 16 (i). And area of PQRS = b 2 So the ratio of area of PQRS and LKXS = b2 b 2 16 = 16 1 = 16:1

2 is an integer is x = 6. Then y = 16 and our triangle has sides of length 6, 16, and 11. Then P = ABC is a triangle that has integer side lengths. BD is an angle bisector, AD = 3, and DC = 8. What is the smallest possible value of the perimeter? 4. In triangle PQR, the median from vertex P is perpendicular to the median from vertex Q. The lengths of sides PR and QR are 6 and 7, respectively. Find the length of side PQ? With Angle Bisector Theorem, we know that. If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then, contradicting the concept of triangle equality (sum of any two sides must be greater than the third side and the length of the third side must be greater than the difference between the two sides). If we use the next lowest values ( and ), the Triangle Inequality is satisfied. Therefore, our answer is = 33 An important observation to make in this problem is that the point G, which is where PE and QS intersect, is the centroid of the triangle. Hence, we know that =PG/GE =QG/GS = 2/1. let s say QG = 2x, GS = x, and let s say PG = 2y, and GE = y. Now, we can use the Pythagorean Theorem on right triangles PGQ and QGE. We get the following two equations: Or Let x be the length of AB and y the length of BC. Note that in this triangle, y/8 = x/3 so that y = 8x/3. Hence P = 11+x+y = 11 + x + 8x/3 The smallest positive integer x such that 8x/3 is an integer is x=3. Then the sides are 3, 8 and 11, but since 3+8 = 11 these cannot be the sides of a triangle. The next largest integer x for which 8x/3 We get: The length of side PQ is Therefore:

3 5. What is the value of AED? DE and AE are the bisectors of ADC and DAB respectively and AB is parallel to CD? 6. How many differently shaped triangles exist in which no two sides are of the same length, each side is of integral unit length and the perimeter of the triangle is less than 14 units? a) 3 b) 4 c) 5 d) 6 e) None of the above Let be a triangle with sides a, b and c, where c is the largest side. a) 80 b) 90 c) 100 d) 120 It is given that the perimeter must be less than 14. a + b + c < 14. Now, sum of two sides is greater than the third side any triangle. c < a + b c + c < a + b + c < 14 c < 7 Hence, we take value of c from 1 to 6, and check if we can find a and b that satisfies (i). These values are tabulated below: c a,b 4 3,2 5 2,4 5 3,4

4 6 2,5 6 3,4 7. In the given triangle, BC = 12cm, DB = 9 cm, CD = 6 cm and BCD = BAC. What is the perimeter of triangle ADC to that of triangle BDC? a) 7/9 b) 8/9 c) 6/9 d) 5/9 The same question has been repeated a) 7/9 b) 8/9 c) 6/9 d) 5/9 Triangle ABC and triangle CBD are similar AC/BC = CD/BD AC/12 = 6/9 AC=8 AB/BC= BC/DB (AD+BD)/BC=BC/DB (AD+9)/12 = 12/9 AD = 7 perimeter of triangle ADC= 21 Perimeter of BDC= 27 Therefore the ratio is 7/9 Triangle ABC and triangle CBD are similar AC/BC = CD/BD AC/12 = 6/9 AC=8 AB/BC= BC/DB (AD+BD)/BC=BC/DB (AD+9)/12 = 12/9 AD = 7 perimeter of triangle ADC= 21 Perimeter of BDC= 27 Therefore the ratio is 7/9 9. If D is the midpoint of side BC of a triangle ABC and AD is the perpendicular to AC, then: a) 3 AC 2 = BC 2 AB 2 b) 3 BC 2 = AC 2 3AB 2 c) 5 AB 2 = BC 2 + AB 2 d) None of the above 8. In the given triangle, BC =12 cm, DB = 9 cm, CD = 6 cm and BCD = BAC, what is the ratio of the perimeter of the triangle ADC to that of triangle BDC?

5 Now, BD = DC, then by Apollonius theorem in ABC, we have, 2AD 2 + 2BD 2 = AB 2 + AC 2 (i) And in ADC, we have, AD 2 = CD 2 - AC 2 (ii) From (i) and (ii), we get, 2(CD 2 - AC 2 ) + 2BD 2 = AB 2 + AC 2 BC 2 AB 2 = 3AC 2 (Beacuse BD 2 = CD 2 = BC2 4 ) 10. In the given figure, PA, QB and RC are each perpendicular to AC, Which one of the following is true? a) x + y = z b) xy = 2z c) 1/x + 1/y = 1/z d) 1/x + 1/y + 1/z = There is a huge circular pillar in a museum. Two people, Ajay and Vijay, go for a visit to the museum, Ajay is standing at a distance of 3 m from point D of the pillar, and Vijay is standing at 9 m from point B of the pillar as shown in the figure. What is the diameter of the circular pillar, if Ajay and Vijay can barely see each other and BD is the diameter of the circular pillar with ABC = 90 0?

6 Applying Pythagoras theorem in OCE CE 2 = (3 + r) 2 r 2 CE 2 = 9 + 6r + r 2 - r 2 CE = 9 + 6r (ii) From (i) and (ii) r(3+2r) 9 = 9 + 6r r 2 (3 + 2r) 2 = 81 (9 + 6r) a) 27 b) 15 c) 9 d) 6 r 2 (3 + 2r) 2 = 243 (3 + 2r) r 2 (3 + 2r) = 243 Now substituting the options. only r = 9 satisfies the given equations. 2 Hence diameter is A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The mayor of the city wants to construct three paths from the corner opposite to the longest side such that these three paths divide the longest side into four equal segments, determine the sum of the squares of the lengths of the three paths? CEO ~ CBA CE Cb = OE AB CE 3+2r = r 9 CE = r(3+2r) 9 (i)

7 b) 39/128 c) 44/128 d) 48/128 e) 64/ ABCD is a rectangle. P,Q and R are the midpoints of BC, CD and DA. The point S lies on the line QR in such a way that SR:QS = 1:3. The ratio of the triangle APS and rectangle ABCD is a) 36/ A right angled triangle ABC is such that ACB = 90 0 and BC = 4cm, M is a point on BC such that AM = 3 cm, now, the perimeter of the triangle AMB is equal to perimeter of triangle ACM. Then the value of BM/ CM is? a) BM/ CM < 1/3

8 b) 1/3 < BM/CM < 1 c) BM/CM > 1 d) None of these of area of triangle DOP to that of triangle AOQ? a) 16:15 b) 256:225 c) 15:16 d) None of these Answer is Option d) Given P ( ACM) = P ( AMB) =) AC + CM + AM = AM + MB + AB =) AC + CM = MB + AB Since, AB > AC (AB is hypotenuse) =) CM > MB =)CM/MB >1 Also, given AM = 3 so CM <3 (AM is hypotenuse) Since, CM +MB = 4, when CM = 3, BM = 1 But Since CM <3 BM/CM >1/3 Option b) is the right answer 16. What is the area of the square inscribed in this right angled triangle that has Pythagorean triplets as 6, 8 and 10 units, that has the base as 8 units? 15. ABCD is a parallelogram such that the opposite sides are parallel and equal. P is a point of line CD and Q is a point on line AB. AQ:QB = 3:1, DP:PC = 4:1. AP and DQ intersect at O. What is the ratio

Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch

Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. in the parallelogram, each two opposite