# math

Problem of the Week

Alternate Dimensions

The four shapes to the right are each drawn
with a horizontal base and a vertical height.
Figure A is a right-angled triangle, Figure B
is an isosceles triangle, Figure C is a square,
and Figure D is a rectangle. The figures are
not drawn to scale.

Using the following clues, determine the measure of the (horizontal) base and the
measure of the (vertical) height of each figure.
1. The measure of the base of Figure A is the same as the measure of the base
of Figure D.
2. The measure of the base of Figure A is one unit less than the measure of the
base of Figure B.
3. The side length of Figure C is the same as the measure of the base of
Figure A.
4. The measure of the height of Figure B is the same as the measure of the
height of Figure A and also the same as the measure of the base of Figure B.
5. The area of Figure C is 9 square units.
6. The total area of all four figures is 38 square units.

1. 👍 0
2. 👎 0
3. 👁 89
1. A = the total area of all four figures

AA = the area of a right-angled triangle

AB = the area of a an isosceles triangle

AC = the area of a an square

AD = the area of a rectangle

LA = the measure of the base of Figure A

LB = the measure of the base of Figure B

LC = the measure of the base of Figure C

LD = the measure of the base of Figure D

LC = the side length of Figure C

hA = measure of the height of Figure A

hB =measure of the height of Figure B

hD = measure of the height of Figure D

Clues:

1.

The measure of the base of Figure A is the same as the measure of the base of Figure D mean:

LA = LD

2.

The measure of the base of Figure A is one unit less than the measure of the base of Figure B mean:

LA = LB - 1

3.

The side length of Figure C is the same as the measure of the base of Figure A mean:

LC = LA

4.

The measure of the height of Figure B is the same as the measure of the height of Figure A and also the same as the measure of the base of Figure B mean:

hB = hA = LB

5.

5. The area of Figure C is 9 square units mean:

LC² = 9

So:

LC² = √9

LC = 3

LC = LA

LA = LC

LA = 3

LA = LB - 1

3 = LB - 1

4 = LB

LB = 4

LA = LD

LD = LA

LD = 3

hB = hA = LB

hA= LB

hA = 4

hB = LB

hB = 4

6.

AA = area of a right-angled triangle

AB = area of a an isosceles triangle

AC = area of a an square

AD = area of a rectangle

A = AA + AB + AC + AD = 38

AA = LA ∙ hA / 2

AA = 3 ∙ 4 / 2 = 12 / 2

AA = 6

AB = LB ∙ hB / 2

AB = 4 ∙ 4 / 2 = 16 / 2

AB = 8

AC = 9

The total area of all four figures:

A = area of a right-angled triangle + area of a an isosceles triangle + area of a an square + area of a rectangle = 38

A = AA + AB + AC + AD = 38

6 + 8 + 9 + 3 hD = 38

23 + 3 hD = 38

Subtract 23 to both sides

3 hD = 15

Divide both sides by 3

hD = 5

Results:

AA = the area of a right-angled triangle = 6

AB = the area of a an isosceles triangle = 8

AC = the area of a an square = 9

AD = the area of a rectangle = 15

LA = the measure of the base of Figure A = 3

LB = the measure of the base of Figure B = 4

LC = the side length of Figure C = 3

LD = the measure of the base of Figure D = 3

hA = measure of the height of Figure A = 4

hB =measure of the height of Figure B = 4

hD = measure of the height of Figure D = 5

1. 👍 0
2. 👎 0
2. One my typo:

It should not be written like this:

So:

LC² = √9

It should be written like this:

So:

LC = √9

1. 👍 0
2. 👎 0

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